5 Marks Questions - Maths STD 10 Questions - Vidyadip

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21 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks

The line segment joining the points A (3, 2) and B (5, 1) is divided at the point P in the ratio 1 : 2 and it lies on the line 3x - 18y + k = 0. Find the value of k.

Answer

Given that, the line segment joining the points 4(3, 2) and 6(5, 1) is divided at the point P in the ratio 1 : 2.
Coordinate of point $\text{P}\equiv\left\{\frac{5(1)+3(2)}{1+2},\frac{1(1)+2(2)}{1+2}\right\}$
$\text{P}\equiv\Big(\frac{5+6}{3},\frac{1+4}{3}\Big)$
$\text{P}\equiv\Big(\frac{11}{3},\frac{5}{3}\Big)$
$\Big[\because$ by section formula for internal ratio $\equiv\Big(\frac{\text{m}_2\text{x}_1+\text{m}_1\text{x}_2}{\text{m}_1+\text{m}_1},\frac{\text{m}_2\text{y}_1+\text{m}_1\text{y}_2}{\text{m}_1+\text{m}_2}\Big)\Big]$
But the point $\text{P}\Big(\frac{11}{3},\frac{5}{3}\Big)$ lies on the line 3x - 18y + k = 0 [given]
$\therefore3\Big(\frac{11}{3}\Big)-18\Big(\frac{5}{3}\Big)+\text{k}=0$
⇒ 11 - 30 + k = 0
⇒ k - 19 = 10
⇒ k = 19
Hence, the required value of k is 19.

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Question 25 Marks

If the point A(2, -4) is equidistant from P(3, 8) and Q(-10, y), find the values of y. Also find distance PQ.

Answer

According to the question,
A(2, -4) is equidistant from P(3, 8) = Q(-10, y) is equidistant from A(2, -4)
i. e., $\text{PA}=\text{QA}$
$\Rightarrow\sqrt{(2-3)^2+(-4-8)^2}$ $=\sqrt{(2-10)^2+(-4-\text{y})^2}$
$\Big[\because$ distance between two points $( x _1, y _1)$ and $( x _2, y _2)$,$\text{ d}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\Big]$
$\Rightarrow\sqrt{(-1)^2+(-12)^2}=\sqrt{(12)^2+(4+\text{y})^2}$
$\Rightarrow\sqrt{1+144}=\sqrt{144+16+\text{y}^2+8\text{y}}$
$\Rightarrow\sqrt{145}=\sqrt{160+\text{y}^2+8\text{y}}$
On squaring both the sides, we get
$145 = 160 + y^2 + 8y$
$y^2 + 8y + 160 - 145 = 0$
$y^2+ 8y + 15 = 0$
$y^2+ 5y + 3y + 15 = 0$
$y(y + 5) + 3(y + 5) = 0$
$(y + 5)(y + 3) = 0$
$If y + 5 = 0, then y = -5$
$If y + 3 = 0, then y = -3$
Now, distance between P(3, 8) and Q(-10, y)
$\text{PQ}=\sqrt{(-10-3)^2+(\text{y}-8)^2}$ [Putting y = -3]
$\Rightarrow\text{PQ}=\sqrt{(13)^2+(-3-8)^2}$
$\Rightarrow\text{PQ}=\sqrt{169+121}$
$\Rightarrow\text{PQ}=\sqrt{290}$
Again, distance between P(3, 8) and (-10, y)
$\text{PQ}=\sqrt{(-13)^2+(-5-8)^2}$ [Putting y = -5]
$\Rightarrow\text{PQ}=\sqrt{169+169}$
$\Rightarrow\text{PQ}=\sqrt{338}$
Hence, the values of y are -3, -5 and corresponding values of PQ are $\sqrt{290}$ and $\sqrt{338}=1342,$ respectively.

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Question 35 Marks

The points A(2, 9), B(a, 5) and C(5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of $\triangle\text{ABC}.$

Answer

Given that, the points A(2, 9), B(a, 5) and C(5, 5) are the vertices of a $\triangle\text{ABC}$ right angled at B.
By Pythagoras theorem, $AC^2 = AB^2 + BC^2 .........(i)$
$\Big[\because$ distance between the points $( x _1, y _1)$ and $( x _2, y _2)$, $\text{d}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\Big]$
$=\sqrt{\text{a}^2+4-4\text{a}+16}$
$=\sqrt{\text{a}^2-4\text{a}+20}$
$\text{BC}=\sqrt{(5-\text{a})^2+(5-5)^2}$
$\text{BC}=\sqrt{(5-\text{a})^2+0}$
$\text{BC}=5-\text{a}$
Put the values of AB, BC, and AC in Eq. (i), we get
$(5)^2=\Big(\sqrt{\text{a}^2-4\text{a}+20}\Big)^2+(5-\text{a})^2$
$\Rightarrow 25 = a^2 - 4a + 20 + 25 + a^2 - 10a$
$\Rightarrow 2a^2- 14a + 20 = 0$
$\Rightarrow a^2 - 7a + 10 = 0$
$\Rightarrow a^2 - 2a - 5a + 10 = 0$
$\Rightarrow a(a - 2) - 5(a - 2) = 0$
$\Rightarrow (a - 2) - (a - 5) = 0$
$\Rightarrow a = 2, 5$
Here, $\text{a}\neq5,$ since at a = 5, the length of BC = 0. It is not Possible because the sides AB, BC and CA form a right angled triangle.
$\Rightarrow a = 2$
Now, the coordinate of A, B and C become (2, 9), (2, 5) and (5, 5), respectively.
$\because\text{Area of }\triangle\text{ ABC}$
$\triangle=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]=0$
$\triangle=\frac{1}{2}[2(5-5)+2(5-9)+5(9-5)]$
$\triangle=\frac{1}{2}[2\times0+2(-4)+5(4)]$
$\triangle=\frac{1}{2}\times12=6$
Hence, the required area of $\triangle\text{ABC}$ is 6 sq units.

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Question 45 Marks

Find the ratio in which the point $\text{P}\Big(\frac{3}{4},\frac{5}{12}\Big)$ divides the line segment joining the points $\text{A}\Big(\frac{1}{2},\frac{3}{2}\Big)$ and B(2, -5).

Answer

Let $\text{P}\Big(\frac{3}{4},\frac{5}{12}\Big)$ divide AB internally in the ratio m : n
Using the section formula, we get
$\Big(\frac{3}{4},\frac{5}{12}\Big)=\Bigg(\frac{2\text{m}-\frac{\text{n}}{2}}{\text{m}+\text{n}},\frac{-5\text{m}+\frac{3}{2}\text{n}}{\text{m}+\text{n}}\Bigg)$
$\Big[\because$ internal section formula, the coordinates of point P divides the line segment joining the point $( x _1, y _1)$ and $( x _2, y _2)$ in the ratio $m_1 : m_2$ internally is $\Big(\frac{\text{m}_2\text{x}_1+\text{m}_1\text{x}_2}{\text{m}_1+\text{m}_1},\frac{\text{m}_2\text{y}_1+\text{m}_1\text{y}_2}{\text{m}_1+\text{m}_2}\Big)\Big]$
On equation, we get
$\frac{3}{4}=\frac{2\text{m}-\frac{\text{n}}{2}}{\text{m}+\text{n}}$ and $\frac{5}{12}=\frac{-5\text{m}+\frac{\text{3}}{2}\text{n}}{\text{m}+\text{n}}$
$\Rightarrow\frac{3}{4}=\frac{4\text{m}-\text{n}}{2(\text{m}+\text{n})}$ and $\frac{5}{12}=\frac{-10\text{m}+3\text{n}}{2(\text{m}+\text{n})}$
$\Rightarrow\frac{3}{4}=\frac{4\text{m}-\text{n}}{\text{m}+\text{n}}$ and $\frac{5}{6}=\frac{-10\text{m}+3\text{n}}{\text{m}+\text{n}}$
⇒ 3m + 2n = 8m - 2n and 5m + 5n = -60m + 18n
⇒ 5n - 5m = 0 and 65m - 13n = 0
⇒ n = m and 13(5m - n) = 0
⇒ n = m and 5m - n = 0
Since, m = n does not satisty.
⇒ 5m - n = 0
⇒ 5m = n
$\frac{\text{m}}{\text{n}}=\frac{1}{5}$
Hence, the required ratio is 1 : 5.

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Question 55 Marks

Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office. What is the extra distance travelled by Ayush in reaching his office? (Assume that all distances covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in km.

Answer

Consider the coordinates of house H(2, 4), bank B(5, 8), school S(13, 14) and office O(13, 14).
Distance $HB^2 = (5 - 2)^2 + (8 - 4)^2$
$\Rightarrow HB^2= 3^2 + 4^2$
$\Rightarrow HB^2 = 9 + 16$
$\Rightarrow HB^2 = 25$
$\Rightarrow HB = 5km$
Distance $BS^2 = (13 - 5)^2 + (14 - 8)^2$
$\Rightarrow BS^2 = 8^2 + 6^2$
$\Rightarrow BS^2 = 64 + 36$
$\Rightarrow BS^2 = 100$
$\Rightarrow BS = 10km$
Distance $SO^2 = (13 - 13)^2 + (26 - 14)^2$
$\Rightarrow SO^2= 0^2 + 12^2$
$\Rightarrow SO^2 = 12^2$
$\Rightarrow SO = 12km$
Total distance travelled by ayush from house to bank to school and then to office
$= HB + BS + SO$
$= 5 + 10 + 12$
$= 27km$
Direct distance from house to office $= HO$
$\Rightarrow HO^2 = (13 - 2)^2 + (26 - 4)^2$
$\Rightarrow HO^2 = (11)^2 + (22)^2$
$\Rightarrow HO^2= 121 + 484$
$\Rightarrow\text{HO}=\sqrt{605}$
HO = 24.6km
So, extra distance travalled by ayush = 27km - 24.6km = 2.4km.
Hence, extra distance travelled by ayush = 2.4km.

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Question 65 Marks

What type of a quadrilateral do the points A(2, -2), B(7, 3), C(11, -1) and D(6, -6) taken in that order, form?

Answer

To find the type of quadrilateral, we find the length of all four sides as well as two diagonals and see whatever condition of quadrilateral is satisfy by these sides as well as diagonals. Now, using distance formula between two points,
$\Big[$ Since, distance between two point $( x _1, y _1)$ and $( x _2, y _2)$ $=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\Big]$

Sides, $\text{AB}=\sqrt{(7-2)^2+(3+2)^2}$
$\text{AB}=\sqrt{(5)^2+(5)^2}$
$\text{AB}=\sqrt{25+25}$
$\text{AB}=\sqrt{50}$
$\text{AB}=5\sqrt{2}$
$\text{BC}=\sqrt{(11-7)^2+(-1-3)^2}$
$\text{BC}=\sqrt{(4)^2+(-4)^2}$
$\text{BC}=\sqrt{16+16}$
$\text{BC}=\sqrt{32}$
$\text{BC}=4\sqrt{2}$
$\text{CD}=\sqrt{(6-11)^2+(-6+1)^2}$
$\text{CD}=\sqrt{(-5)^2+(-5)^2}$
$\text{CD}=\sqrt{25+25}$
$\text{CD}=\sqrt{50}$
$\text{CD}=5\sqrt{2}$
and $\text{DA}=\sqrt{(2-6)^2+(-2+6)^2}$
$\text{DA}=\sqrt{16+16}$
$\text{DA}=\sqrt{32}$
$\text{DA}=4\sqrt{2}$
Diagonals, $\text{AC}=\sqrt{(11-2)^2+(-1+2)^2}$
$\text{AC}=\sqrt{(9)^2+(1)^2}$
$\text{AC}=\sqrt{81+1}$
$\text{AC}=\sqrt{82}$
and $\text{BD}=\sqrt{(6-7)^2+(-6-3)^2}$
$\text{BD}=\sqrt{(-1)^2+(-9)^2}$
$\text{BD}=\sqrt{1+81}$
$\text{BD}=\sqrt{82}$
Here, we see that the sides AB = CD and BC = DA
Also, diagonals are equal i. e., AC = BD
Which shows the quadrilateral is a rectangle.

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Question 75 Marks

In what ratio does the x–axis divide the line segment joining the points (-4, -6) and (-1, 7)? Find the coordinates of the point of division.

Answer

Let the required ratio be $\lambda:1.$ So, the coordinates of the point M of division A (- 4, -6) and B(-1,7) are
$\Big(\frac{\lambda(-1)+1(-4)}{\lambda+1},\frac{\lambda(7)+1(-6)}{\lambda+1}\Big)$$=\Big(\frac{\lambda-4}{\lambda+1},\frac{7\lambda-6}{\lambda+1}\Big)$
But according to the question, line segment joining A(-4, -6) and B(-1, 7) is divided by the x-axis. So, y-coordinate must be zero.
$\therefore\frac{7\lambda-6}{\lambda+1}$
$\Rightarrow7\lambda-6=0$
$\therefore\lambda=\frac{6}{7}$
So, the required ratio is 6 : 7 and the point of division M is
$\left\{\frac{-\frac{6}{7}-4}{\frac{6}{7}+1},\frac{7\times\frac{6}{7}-6}{\frac{6}{7}+1}\right\}$
i. e., $\Bigg(\frac{\frac{-34}{7}}{\frac{13}{7}},\frac{6-6}{\frac{13}{7}}\Bigg)$
i. e., $\Big(\frac{-34}{13},0\Big)$
Hence, the required point of division is $\Big(\frac{-34}{13},0\Big).$

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Question 85 Marks

The points $A(x_1, y_1), B(x_2, y_2)$ and $C(x_3, y_3)$ are the vertices of $\triangle\text{ABC}.$
Find the coordinates of points Q and R on medians BE and CF, respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.

Answer

Midian BE meets the sides AC at its mid-point E.

$\therefore\text{Coordinates of E are}\Big(\frac{\text{x}_1+\text{x}_3}{2},\frac{\text{y}_1+\text{y}_3}{2}\Big).$

Now, the corrdinates of Q such that BE is median and BQ : QE = 2 : 1 are given by

$\text{x}=\frac{2\Big(\frac{\text{x}_2+\text{x}_3}{2}\Big)+1(\text{x}_2)}{2+1},$ $\text{y}=\frac{2\Big(\frac{\text{y}_2+\text{y}_3}{2}\Big)+1(\text{y}_2)}{2+1}$

$\Rightarrow\text{x}=\frac{\text{x}_1+\text{x}_3+\text{x}_2}{3},$ $\Rightarrow\text{y}=\frac{\text{y}_1+\text{y}_3+\text{y}_2}{3}$

The coordinates of point Q on medianj BE such at QB : QE = 2 : 1 are $\Big(\frac{\text{x}_1+\text{x}_2+\text{x}_2}{3},\frac{\text{y}_1+\text{y}_2+\text{y}_3}{3}\Big).$

Median CF meets the side AB at its mid-point F.

$\therefore\text{Coordinates of F are}\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big).$

Now, the coordinates of R such that CF is median and CR : RF = 2 : 1 are given by

$\text{x}=\frac{1(\text{x}_3)+2\Big(\frac{\text{x}_1+\text{x}_2}{2}\Big)}{1+2},$ $\text{y}=\frac{1(\text{y}_3)+2\Big(\frac{\text{y}_1+\text{y}_2}{2}\Big)}{1+2}$

$\Rightarrow\text{x}=\frac{\text{x}_3+\text{x}_1+\text{x}_2}{3},$ $\text{y}=\frac{\text{y}_3+\text{y}_1+\text{y}_2}{3}$

So, the coordinates point R on the median CF such that CR : RF = 2 : 1 are $\Big(\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3},\frac{\text{y}_1+\text{y}_2+\text{y}_3}{3}\Big)$

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Question 95 Marks

If P(9a, -2, -b) divides line segment joining A(3a + 1, -3) and B(8a, 5) in the ratio 3 : 1, find the values of a and b.

Answer

Let P(9a - 2, -b) divides AS internally in the ratio 3 : 1.
By section formula,
By section fotmula, the coordinates of point P are given as:
$\Big(\frac{\text{m}_2\text{x}_1+\text{m}_1\text{x}_2}{\text{m}_1+\text{m}_1},\frac{\text{m}_2\text{y}_1+\text{m}_1\text{y}_2}{\text{m}_1+\text{m}_2}\Big)$
$=\Big(\frac{3(8\text{a})+1(3\text{a}+1)}{3+1}.\frac{3(5)+1(-3)}{3+1}\Big)$
$\Rightarrow9\text{a}-2=\frac{3(8\text{a})+1(3\text{a+1})}{3+1}$
and $-\text{b}=\frac{3(5)+1(-3)}{3+1}$
$\Rightarrow9\text{a}-2=\frac{24\text{a}+3\text{a}+1}{4}$
and $-\text{b}=\frac{15-3}{4}$
$\Rightarrow9\text{a}-2=\frac{27\text{a}+1}{4}$
and $-\text{b}=\frac{12}{4}$
$\Rightarrow36\text{a}-8=27\text{a}+1$
and $\text{b}=-3$
Hence proved a = 1 and b = -3.

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Question 105 Marks

Students of a school are standing in rows and columns in their playground for a drill practice. A, B, C and D are the positions of four students as shown in Is it possible to place Jaspal in the drill in such a way that he is equidistant from each of the four students A, B, C and D? If so, what should be his position?

Answer

Corrdinates of A, B, C and D from graph are A(3, 5), B(7, 9), C(11, 5) and D(7, 1).
To find the shape of $\Box\text{ABCD},$
$\Rightarrow AB^2 = (7 - 3)^2 + (9 - 5)^2$
$\Rightarrow AB^2 = 4^2 + 4^2$
$\Rightarrow AB^2= 4^2(1 + 1)$
$\Rightarrow\text{AB}=4\sqrt{2}\text{ units}$
$\therefore\text{AB}=\text{BC}=\text{CD}=\text{DA}=4\sqrt{2}\text{ units}$
So, ABCD will be either square or rhombus.
Now, Diagonal $\text{AC}=\sqrt{(11-3)^2+(5-5)^2}$
$\Rightarrow\text{AC}=\sqrt{(8)^2+(0)^2}$
$\Rightarrow\text{AC}=8\text{ units}$
$\therefore$ Diagonal AC = Diagonal BD
So, the given quadrilateral ABCD is a square. the point which is equidistant from point A, B, C, D of a square ABCD will be at the intersecting point of diagonals and diagonals bisect each other.
Hence, the requored point O equidistant from A, B, C, D is mid point of any diagonal $=\Big(\frac{7+7}{2},\frac{9+1}{2}\Big)=\Big(\frac{14}{2},\frac{10}{2}\Big)=(7,5).$
Hence, the required point is (7, 5).

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Question 115 Marks

Find the ratio in which the line 2x + 3y - 5 = 0 divides the line segment joining the points (8, -9) and (2, 1). Also find the coordinates of the point of division.

Answer

Let the line 2x + 3y - 5 = 0 divides the line segment joining the points A (8, -9) and B (2, 1) in the ratio $\lambda:1$ at point P.
$\therefore$ Coordinates of $\text{P}\equiv\left\{\frac{2\lambda+8}{\lambda+1},\frac{\lambda-9}{\lambda+1}\right\}$
$\bigg[\because$ internal division $\left\{\frac{\text{m}_2\text{x}_1+\text{m}_1\text{x}_2}{\text{m}_1+\text{m}_1},\frac{\text{m}_2\text{y}_1+\text{m}_1\text{y}_2}{\text{m}_1+\text{m}_2}\Big)\right\}\bigg]$
But P lies on $2\text{x} + 3\text{y} - 5 = 0$
$\therefore2\Big(\frac{2\lambda+8}{\lambda+1}\Big)+3\Big(\frac{\lambda-9}{\lambda+1}\Big)-5=0$
$\Rightarrow2(2\lambda+8)+3(\lambda-9)-5(\lambda+1)=0$
$\Rightarrow4\lambda+16+3\lambda-27-5\lambda-5=0$
$\Rightarrow2\lambda-16=0$
$\Rightarrow\lambda=8$
$\Rightarrow\lambda:1$
$\Rightarrow8:1$
So, the point P divides the lies the line in the ratio 8 : 1
$\therefore$ Point of division $\text{P}\equiv\left\{\frac{2(8)+8}{8+1},\frac{8-9}{8+1}\right\}$
$\text{P}\equiv\Big(\frac{16+8}{9},-\frac{1}{9}\Big)$
$\text{P}\equiv\Big(\frac{24}{9},\frac{-1}{9}\Big)$
$\text{P}\equiv\Big(\frac{8}{3},\frac{-1}{9}\Big)$
Hence, the requied point of division is $\Big(\frac{8}{3},\frac{-1}{9}\Big).$

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Question 125 Marks

If the points A(1, -2), B(2, 3), C(a, 2) and D(- 4, -3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.

Answer

As ABCD is a parallelogram and diagonals of parallelogram bisect each other.

The mid points of diagonals of parallelogram will coincide i. e.,
Mid-point of diagonal AC = Mid-point of diagonal BD
$\Rightarrow\Big(\frac{1+\text{a}}{2},\frac{-2+2}{2}\Big)=\Big(\frac{-4+2}{2},\frac{-3+3}{2}\Big)$
$\Rightarrow\Big(\frac{1+\text{a}}{2},0\Big)=\Big(\frac{-2}{2},0\Big)$
$\Rightarrow\frac{1+\text{a}}{2}=\frac{-2}{2}$
$\Rightarrow\text{a}=-2-1$
$\Rightarrow\text{a}=-3$
Hence, the value of a is -3.
$\text{Area of }\triangle\text{ABD}=\frac{1}{2}\text{base}\times\text{altitude}$
$\Rightarrow\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]=\frac{1}{2}\text{AB}\times\text{h}$
$\Rightarrow\frac{1}{2}[1\{3-(-3)\}+2\{-3-(-2)\}-4(-2-3)]$ $=-\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$\Rightarrow\frac{1}{2}[(3+3)+2(-3+2)-4(-5)]$ $=\frac{\text{h}}{2}\sqrt{(2-1)^2+(3+2)^2}$
$\Rightarrow\frac{1}{2}[6+2(-1)+20]$ $=\frac{\text{h}}{2}\sqrt{(1)^2+(5)^2}$
$\Rightarrow\frac{1}{2}[6-2+20]$ $=\frac{\text{h}}{2}\sqrt{1+25}$
$\Rightarrow\frac{1}{2}[26-2]$ $=\frac{\text{h}}{2}\sqrt{26}$
$\Rightarrow\text{h}\sqrt{26}$ $=24$
$\Rightarrow\text{h}=\frac{24}{\sqrt{26}}\times\frac{\sqrt{26}}{\sqrt{26}}$
$\Rightarrow\text{h}=\frac{24\sqrt{26}}{26}$
$\Rightarrow\text{h}=\frac{12}{13}\sqrt{26}\text{ units}$
Hence, the perpendicular distance between parallel sides AB and CD is $\frac{12\sqrt{26}}{13}$ units.

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Question 135 Marks

Find the coordinates of the point R on the line segment joining the points P(-1, 3) and Q(2, 5) such that $\text{PR}=\frac{3}{5}\text{PQ}.$

Answer

According to the question,

Given that, $\text{PR}=\frac{3}{5}\text{PQ}$
$\Rightarrow\frac{\text{PQ}}{\text{PR}}=\frac{5}{3}$
$\Rightarrow\frac{\text{PR + RQ}}{\text{PR}}=\frac{5}{3}$
$\Rightarrow1+\frac{\text{RQ}}{\text{PR}}=\frac{5}{3}$
$\Rightarrow\frac{\text{PQ}}{\text{PR}}=\frac{5}{3}+1$
$\Rightarrow\frac{\text{PQ}}{\text{PR}}=\frac{2}{3}$
$\therefore\text{RQ}:\text{PR}=2:3$
or $\text{PR}:\text{RQ}=3:2$
Suppose, R(x, y) be the point which divides the line segment joining the points P(-1, 3) and Q(2, 5) in the ratio 3 : 2.
$\therefore(\text{x, y})=\left\{\frac{3(2)+2(-1)}{3+2},\frac{ 3(5)+2(3)}{3+2}\right\}$
$\bigg[\because$ by internal section formula, $\left\{\frac{\text{m}_2\text{x}_1+\text{m}_1\text{x}_2}{\text{m}_1+\text{m}_1},\frac{\text{m}_2\text{y}_1+\text{m}_1\text{y}_2}{\text{m}_1+\text{m}_2}\Big)\right\}\bigg]$
$\Rightarrow(\text{x, y})=\Big(\frac{6-5}{5},\frac{15+6}{5}\Big)$
$\Rightarrow(\text{x, y})=\Big(\frac{4}{5},\frac{21}{5}\Big)$
Hence, the required coordinates of the point R is $\Big(\frac{4}{5},\frac{21}{5}\Big).$

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Question 145 Marks

Find the coordinates of the point Q on the x–axis which lies on the perpendicular bisector of the line segment joining the points A(-5, -2) and B(4, -2). Name the type of triangle formed by the points Q, A and B.

Answer

Firstly, we plot the points of the line segment on the paper and join them.

We know that, the perpendicular bisector of the line segment AB bisect the segment AB,
i.e., perpendicular bisector of the line segment AB passes through the mid-point of AB.
$\therefore\text{Mid-point of AB}=\Big(\frac{-5+4}{2},\frac{-2-2}{2}\Big)$
$\Rightarrow\text{R}=\Big(\frac{-1}{2},-2\Big)$
$\Big[\because$mid-point of a line segment passes throught the points $( x _1, y _1)$ and $( x _2, y _2)$ is $\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big)\Big]$
Now, we draw a staight line on paper passes through the mid-point R. we see that perpendicular bisector cuts the x-axis at the point $\text{Q}\Big(-\frac{1}{2},0\Big)$
Hence, the required coordinates of $\text{Q}=\Big(-\frac{1}{2},0\Big)$
Alternate Answer
To find the coordinates of the point of 0 on the X-axis. We find the equation of perpendicular bisector of the line segment AS. Now, slope of line segment AB,
Let $\text{m}_1=\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1}$
$\Rightarrow\text{m}_1=\frac{-2-(-2)}{4-(-5)}$
$\Rightarrow\text{m}_1=\frac{-2+2}{4+5}$
$\Rightarrow\text{m}_1=\frac{0}{9}$
$\Rightarrow\text{m}_1=0$
Let the slope of perpendicular bisector of line segment is $m_2$.
Since, perpendicular bisector is perpendicular to the line segment AB
By perpendicularity condition of two lines,
$\text{m}_1:\text{m}_2=-1$
$\Rightarrow\text{m}_2=\frac{-1}{\text{m}_1}=\frac{-1}{0}$
$\Rightarrow\text{m}_2=\infty$
Also, we know that, the perpendicular bisector is always passes throught the mid-point of the line segment.
$\therefore\text{Mid-point}=\Big(\frac{-5+4}{2},\frac{-2-2}{2}\Big)$
$\Big(\frac{-1}{2},-2\Big)$
$\Big[\because\text{Mid-point}=\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big)\Big]$
To find the equation of perpendicular bisector of line segment, we find the slope and a point through which perpendicular bisector is pass
Now, equation of perpendicular bisector having slope $\infty$ and passing through the point $\Big(\frac{-1}{2},-2\Big)$ is,
$[\because(\text{y}-\text{y}_1)=\text{m}_2(\text{x}-\text{x}_1)]$
$\Rightarrow(\text{y}-2)=\infty\Big(\text{x}+\frac{1}{2}\Big)$
$\Rightarrow\frac{\text{y}+2}{\text{x}+\frac{1}{2}}=\infty=\frac{1}{2}$
$\Rightarrow\text{x}+\frac{1}{2}=0$
$\therefore\text{x}=\frac{-1}{2}$
So, the coordinates of the point Q is $\Big(\frac{-1}{2},0\Big)$ on the x-axis which lies on the perpendicular bisector of the line segment joining the point AB.
To know the type of triangle formed by the points Q, A and B. We find the length of all three sides and see whatever condition of triangle is satisfy by these sides.

Now, using distance formula between two points,
$\text{AB}=\sqrt{(4+5)^2+(-2+2)^2}$
$\text{AB}=\sqrt{(9)^2+0}$
$\text{AB}=9$
$\Big[\because$ distance between two points $( x _1, y _1)$ and $( x _2, y _2)$ $=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\Big]$
$\text{BQ}=\sqrt{\Big(\frac{-1}{2}-4\Big)^2+(0+2)^2}$
$\text{BQ}=\sqrt{\Big(\frac{-9}{2}\Big)^2+(2)^2}$
$\text{BQ}=\sqrt{\frac{81}{4}+4}$
$\text{BQ}=\sqrt{\frac{97}{4}}$
$\text{BQ}=\sqrt{\frac{97}{2}}$
$\text{QA}=\sqrt{\Big(-5+\frac{1}{2}\Big)+(-2-0)^2}$
$\text{QA}=\sqrt{\Big(\frac{-9}{2}\Big)^2+(2)^2}$
$\text{QA}=\sqrt{\frac{81}{4}+4}$
$\text{QA}=\sqrt{\frac{97}{4}}$
$\text{QA}=\sqrt{\frac{97}{2}}$
We see that, $\text{BQ}=\text{QA}\neq\text{AB}$
Which shows that the triangle formed by the point Q, A and 6 an isosceles.

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Question 155 Marks

Find the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k - 1, 5k) are collinear.

Answer

We know that, if three points are collinear, then the area of triangle formed by these points is zero.
Since, the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k - 1, 5k) are collinear.
Then, area of $\triangle\text{ABC}=0$
$\Rightarrow\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]=0$
Here, $\text{x}_1=\text{k}+1,\text{ x}_2=3\text{k},\text{ x}_3=5\text{k}-1$
and $\text{y}_1=2\text{k}_1,\text{ y}_2=2\text{k}+3,\text{ y}_3=5\text{k}$
$\Rightarrow\frac{1}{2}\big[(\text{k+1})(2\text{k}+3-5\text{k})+3\text{k}(5\text{k}-2\text{k})+(5\text{k}-1)(2\text{k}-(2\text{k}+3))\big]=0$
$\Rightarrow\frac{1}{2}\big[(\text{k}+1)(-3\text{k}+3)+3\text{k}(3\text{k})+(5\text{k}-1)(2\text{k}-2\text{k}-3)\big]=0$
$\Rightarrow\frac{1}{2}[-3\text{k}^2+3\text{k}-3\text{k}+3+9\text{k}^2-15\text{k}+3]=0$
$\Rightarrow\frac{1}{2}(6\text{k}^2-15\text{k}+6)=0$ [multiply by 2]
$\Rightarrow6\text{k}^2-15\text{k}+6=0$ [by factorisation method]
$\Rightarrow2\text{k}^2-5\text{k}+2=0$ [divide by 3]
$\Rightarrow2\text{k}^2-4\text{k}-\text{k}+2=0$
$\Rightarrow2\text{k}(\text{k}-2)-1(\text{k}-2)=0$
$\Rightarrow(\text{k}-2)(2\text{k}-1)=0$
If $\text{ k}-2, $ then $\text{ k}=2$
If $ 2\text{k}-1=0,$ then $\text{ k}=\frac{1}{2}$
$\therefore\text{ k}=2,\frac{1}{2}$
Hence, the required values of k are 2 and $\frac{1}{2}.$

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Question 165 Marks

If $\text{D}\Big(\frac{-1}{2},\frac{5}{2}\Big),\text{ E}(7,3)$ and $\text{F}=\Big(\frac{7}{2},\frac{7}{2}\Big)$ are the mid-point of sides of $\triangle\text{ABC},$ find the area of the $\triangle\text{ABC}.$

Answer

Let $\text{A}\equiv(\text{x}_1,\text{ y}_1),\text{B}\equiv(\text{x}_2,\text{ y}_2)$ and $\text{C}\equiv(\text{x}_3,\text{ y}_3)$ are the vertices of the $\triangle\text{ABC}.$
Gives, $\text{D}\Big(\frac{-1}{2},\frac{5}{2}\Big),\text{ E}(7,3)$ and $\text{F}=\Big(\frac{7}{2},\frac{7}{2}\Big)$ be the mid-point of the sides BC, CA and AB respectively.
Since, $\text{D}\Big(\frac{-1}{2},\frac{5}{2}\Big)$ is the mid-point of BC.
$\therefore\frac{\text{x}_2+\text{x}_3}{2}=-\frac{1}{2}$
$\Big[\because$ mid-point of a line segment passes throught the points $( x _1, y _1)$ and $( x _2, y _2)$ is $\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big)\Big]$
and $\frac{\text{y}_2+\text{y}_3}{2}=\frac{5}{2}$
$\Rightarrow\text{x}_2+\text{x}_3=-1\ .....(\text{i})$
and $\text{y}_2+\text{y}_3=5\ ......(\text{ii})$
As E(7, 3) is the mid-point of CA
$\therefore\frac{\text{x}_3+\text{x}_1}{2}=7$
and $\frac{\text{y}_3+\text{y}_1}{2}=3$
$\Rightarrow\text{x}_3+\text{x}_1=14\ .....(\text{iii})$
and $\text{y}_3+\text{y}_1=6\ .....(\text{iv})$
Also $\text{F}=\Big(\frac{7}{2},\frac{7}{2}\Big)$ is the mid-point of AB.
$\therefore\frac{\text{x}_1+\text{x}_2}{2}=\frac{7}{2}$
and $\frac{\text{y}_1+\text{y}_2}{2}=\frac{7}{2}$
$\Rightarrow\text{x}_1+\text{x}_2=7\ ......(\text{v})$
and $\text{y}_1+\text{y}_2=7\ .....(\text{iv})$
On adding Eqs. (i), (iii) and (v), we get
$2(x_1 + x_2 + x_3) = 20$
$x_1 + x_2 + x_3 = 10$
On subtracting Eqs. (i), (iii) and (v) from Eq. (vii) respectively, we get
$x_1 = 11, x_2 = -4, x_3 = 3$
On addting Eqs. (ii), (iv) and (vi), we get
$2(y_1 + y_2 + y_3) = 18$
$y_1 + y_2 + y_3 = 9$
On subtracting Eq. (ii), (iv) and (vi) from Eq. (vii) respectively, we get
$y_1 = 4, y_2 = 3, y_3 = 2$
Hence, the vertices of $\triangle\text{ABC}$ are A(11, 4), B(-4, 3) and C(3, 2)
$\because\text{Area of }\triangle\text{ ABC}$
$\triangle=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]=0$
$\triangle=\frac{1}{2}[11(3-2)+(-4)(2-4)+3(4-3)]$
$\triangle=\frac{1}{2}=[11\times1+(-4)(-2)+3(1)]$
$\triangle=\frac{1}{2}(11+8+3)$
$\triangle=\frac{22}{2}=11$
Required area of $\triangle\text{ABC}=11.$

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Question 175 Marks

A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of $\triangle\text{ADE}.$

Answer

ABCD is a paralleiogram so
[Mid point of diagonal BD] = [Mid point of diagonal AC]
$\therefore\text{Mid point of BD}=\Big(\frac{\text{x}_4+8}{2},\frac{\text{y}_4+2}{2}\Big)$
$\text{and Mid point of AC}=\Big(\frac{6+9}{2},\frac{1+4}{2}\Big)$
$\Rightarrow\frac{\text{x}_4+8}{2}=\frac{15}{2}$ and $\frac{\text{y}_4+2}{2}=\frac{5}{2}$
$\Rightarrow\text{x}_4=15-8$ and $\text{y}_4=5-2$
$\Rightarrow\text{x}_4=7$ and $\text{y}_4=3$
$\therefore\text{D}=(7,3)$
$\text{Mid point of DC is}=\text{E}\Big(\frac{\text{x}_4+9}{2},\frac{\text{y}_4+4}{2}\Big)$
$=\text{E}\Big(\frac{7+9}{2},\frac{3+4}{2}\Big)$
$=\text{E}\Big(\frac{16}{9},\frac{7}{2}\Big)$
$=\text{E}\Big(8,\frac{7}{2}\Big)$
Now, Area of $\triangle\text{ADE}=\frac{1}{2}\big[6\big(3-\frac{7}{2}\big)+7\big(\frac{7}{2}-1\big)+8(1-3)\big]$
$=\frac{1}{2}\Big[6\Big(\frac{-1}{2}\Big)+7\Big(\frac{5}{2}\Big)+8(-2)\Big]$
$=\frac{1}{2}\Big(3+\frac{35}{2}-16\Big)$
$=\frac{1}{2}\Big(\frac{-6+35-32}{2}\Big)$
$=\frac{1}{2}\times\frac{(-3)}{2}$
$=\frac{-3}{4}\text{sq. units}$
$=\frac{3}{4}\text{sq. units}$ [In magnitude]
Hence, the area of $\triangle\text{ADE}\text{ is }\frac{3}{4}\text{sq. units}.$

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Question 185 Marks

The centre of a circle is (2a, a - 7). Find the values of a if the circle passes through the point (11, -9) and has diameter $10\sqrt{2}$ units.

Answer

By given condition,

Distance between the centre C(2a, a - 7) and the point P( 11, -9), which lie on the circle = Radius of circle
$\Big[\because$ distance between the points $( x _1, y _1)$ and $( x _2, y _2)$, $\text{d}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\Big]$
Radius of circle $=\sqrt{(11-2\text{a}^2)+(-9-\text{a}+7)^2}$
Given that length of diameter $=10\sqrt{2}$
$\therefore\text{Length of radius}=\frac{\text{Length of diameter}}{2}$
$=\frac{10\sqrt{2}}{2}=5\sqrt{2}$
Put this value in Eq. (i), we get
$5\sqrt{2}=\sqrt{(11-2\text{a})^2+(-2-\text{a})^2}$
Squaring on both sides, we get
$\Rightarrow 50 = (11 - 2a)^2 + (2 + a)^2$
$\Rightarrow 50 = 121 + 4a^2 - 44a + 4 + a^2+ 4a$
$\Rightarrow 5a^2 - 40a + 75 = 0$
$\Rightarrow a^2 - 8a + 15 = 0$
$\Rightarrow a^2 - 5a - 3a + 15 = 0$
$\Rightarrow a(a - 5) - 3(a - 5) = 0$
$\Rightarrow (a - 5)(a - 5) = 0$
$\therefore$ a = 3, 5
Hence, the required values of a are 5 and 3.

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Question 195 Marks

If (- 4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.

Answer

Let A(-4, 3), B(4, 3) and C(x, y) are the three vertices of $\triangle\text{ABC}.$
As the triangle is equileral,
So $AC = BC = AB$
$AC^2 = BC^2 = AB^2 .......(i)$
Now, $AB^2 = (4 + 4)^2 + (3 - 3)^2$
$AB^2 = (8)^2$
$AB = 64$
$AB = 8 units ........(ii)$
$AC^2 = (x + 4)^2 + (y - 3)^2$
$AC^2 = (x)^2 + (4)^2 + 2(x)(4) +(y)^2 + (3)^2 - 2(y)(3)$
$AC^2= x^2 + y^2 + 8x - 6y + 16 + 9$
$AC^2 = x^2 + y^2 + 8x - 6y + 25 ........(iii)$
$BC^2= (x - 4)^2+ (y - 3)^2$
$BC^2 = (x)^2 + (4)^2 - 2(x)(4) + (y)^2 + (3)^2 - 2(y)(3)$
$BC^2= x^2 + y^2 - 8x - 6y + 16 + 9$
$BC^2 = x^2 + y^2 - 8x - 6y + 25 ..........(iv)$
Now, $AC^2 = AB^2$[From (i)]
$x^2 + y^2+ 8x - 6y + 25 = 64 [From (iii), (ii)]$
$x^2 + y^2 + 8x - 6y = 64 - 25$
$x^2 + y^2+ 8x - 6y = 39 ......(v)$
Again $BC^2 = AB^2 $[From (i)]
$x^2 + y^2 - 8x - 6y + 25 = 64 [From (ii), (iv)]$
$x^2 + y^2 - 8x - 6y = 64 - 25$
$x^2 + y^2 - 8x - 6y = 39 .....(vi)$
Subtracting (v) from (vi), we have

$\Rightarrow x = 0$
Putting x = 0 in (v), we have
$(0)^2 + y^2 + 8(0) - 6y = 39$
$\Rightarrow y^2- 6y - 39 = 0$
$\Rightarrow D = b^2 - 4ac (a = 1, b = -6, c = -39)$
$\Rightarrow D = (-6)^2 - 4(1)(-39)$
$\Rightarrow D = 36 + 156$
$\Rightarrow D = 192$
$\Rightarrow\sqrt{\text{D}}=\sqrt{2\times2\times2\times2\times2\times2\times3}$
$\Rightarrow\sqrt{\text{D}}=8\sqrt{3}$
$\therefore\text{y}=\frac{-\text{b}\pm\sqrt{\text{D}}}{2\text{a}}$
$\Rightarrow\text{y}=\frac{6\pm8\sqrt{3}}{2\times1}$
$\Rightarrow\text{y}=\frac{2(3\pm4\sqrt{3})}{2}$
$\Rightarrow\text{y}_1=3+4\sqrt{3}$
and $\text{y}_2=3-4\sqrt{3}$
Hence, the third vertex of $\triangle\text{ABC}$ maybe $\text{C}(0,3+4\sqrt{3})$ and $\text{C}(0,3-4\sqrt{3})$

Now, $\text{C}(0,3+4\sqrt{3})$
$= C(0, 3 + 4 \times 1.732)$
$= C(0, 3 + 6.9)$
$= C(0, 9.9)$
and $\text{C}'(0,3-4\sqrt{3})$
$= C'(0, 3 - 4 × 1.732)$
$= C'(0, 3 - 6.9)$
$= C'(0, -3.09)$
So, the required point so that origin lies inside it is $(0,3-4\sqrt{3}).$

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Question 205 Marks

If (a, b) is the mid-point of the line segment joining the points A(10, -6) and B(k, 4) and a - 2b = 18, find the value of k and the distance AB.

Answer

Since, (a, b) is the mid-point of line segment AB.
$\therefore(\text{a},\text{b})=\Big(\frac{10+\text{k}}{2},\frac{-6+4}{2}\Big)$
$\Big[$ Since, mid-point of a line segment having point $( x _1, y _1)$ and $( x _2, y _2)$ $=\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big)\Big]$
$\Rightarrow(\text{a, b})=\Big(\frac{10+\text{k}}{2},-1\big)$
Now, equating coordinates on both sides, we get
$\therefore\text{a}=\frac{10+\text{k}}{2}\ .....(\text{i})$
and b = -1
Given a - 2b = 18
a - 2(-1) = 18 From Eq. (ii),
⇒ a + 2 = 18
⇒ a = 16
$\Rightarrow16=\frac{10+\text{k}}{2}$ From Eq. (i)
⇒ 32 = 10 + k
⇒ k = 22
Hence, the required value of is 22.
⇒ k = 22
A = (10, -6)
B = (22, 4)
Now, distance between A(10, -6) and (22, 4)
$\text{AB}=\sqrt{(22-10)^2+(4+6)^2}$
$\Big[\because$ distance between the points $( x _1, y _1)$ and $( x _2, y _2)$, $\text{d}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\Big]$
$\text{AB}=\sqrt{(12)^2+(10)^2}$
$\text{AB}=\sqrt{144+100}$
$\text{AB}=\sqrt{244}$
$\text{AB}=2\sqrt{61}$
Hence, the required distance of AB is $2\sqrt{61}.$

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Question 215 Marks

Find a point which is equidistant from the points A(-5, 4) and B(-1, 6)? How many such points are there?

Answer

Let P (h, k) be the point which is equidistant from the points A (- 5, 4) and B (-1, 6).
$\Big[\because$ distance between two point $=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\Big]$
$\therefore$ PA = PB
$\Rightarrow (PA)^2 = (PB)^2$
$\Rightarrow (-5 - h)^2 + (4 - k)^2 = (-1 - h)^2 + (6 - k)^2$
$\Rightarrow 25 + h^2+ 10h + 16 + k^2 - 8k = 1 + h^2 + 2h + 36 + k^2 - 12k$
$\Rightarrow 25 + 10h + 16 - 8k = 1 + 2h + 36 - 12k$
$\Rightarrow 8h + 4k + 4 = 0$
$\Rightarrow 2h + k + 1 = 0$
Mid-point $\text{AB}=\Big(\frac{-5-1}{2},\frac{4+6}{2}\Big)=(-3, 5)$
$\Big[\because\text{mid-point}=\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big)\Big]$
At point (-3, 5), from EQ. (i).
$\Rightarrow 2h + k = 2(-3) + 5$
$\Rightarrow 2h + k = -6 + 5$
$\Rightarrow 2h + k = -1$
$\Rightarrow 2h + k +1 = 0$
So, the mid-point of AB satisfy the Eq. (i). Hence, infinite number of points, in fact all points which are solution of the equation 2h + k + 1 = 0, are equidistant from the points A and B. Replacing h, kby, x, y in above equation, we have 2 x + y + 1= 0.

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