Question 12 Marks
Find the ratio in which the line segment joining the points A(3,8) and B(-9, 3)is divided by the Y-axis.
AnswerA point P(x,y) divides the line formed by points (a,b) and (c,d) in the ratio of m:n, then the coordinates of the point P is given by
$x =\frac{ an + cm }{ m + n } \text { and } y =\frac{ bn + dm }{ m + n }$
On y axis x coordinate = 0
Let the y-axis divide AB is ratio 1:k.
the points A(3,8) and B(-9, 3)is divided by the Y-axis.
For x coordinate
$0=\frac{3 k -9}{ k +1}$
Solving for k we get
3k - 9 = 0
3 k = 9
k = 3
Hence the y axis divides the given point in the ratio 3:1
View full question & answer→Question 22 Marks
Find k, if PQ || RS and P(2, 4), Q (3, 6), R(3, 1), S(5, k) .
Answertwo lines are said to be parallel if their slopes are equal
If PQ||RS then their slopes must be equal
Slope of $PQ =\frac{6-4}{3-2}=2$
Slope pf RS $=\frac{k-1}{5-3}$
As their slopes are equal, we get
$2=\frac{( k -1)}{2}$
Simplifying
$k-1=4 $
$k=5$
View full question & answer→Question 32 Marks
Find k, if B(k, -5), C (1, 2) and slope of the line is 7.
AnswerSlope m of a line passing through two points A(a,b) and B(c,d) ig given by
$m=\frac{d-b}{c-a}$
In the given question
$7=\frac{2-(-5)}{1-k}$
Simplifying
7-7k = 7
k = 0
View full question & answer→Question 42 Marks
Find k, if R(1, -1), S (-2, k) and slope of line RS is -2.
AnswerSlope m of a line passing through two points A(a,b) and B(c,d) ig given by
$m =\frac{ d - b }{ c - a }$
In the given question
$-2=\frac{ k -(-1)}{-2-1}$
Simplifying
6 = k + 1
K = 5
View full question & answer→Question 52 Marks
A(h, -6), B(2, 3) and C(-6, k) are the co-ordinates of vertices of a trianglewhose centroid is G (1, 5). Find h and k.
AnswerThe coordinates of the centroid (x,y) od a triangle formed by points (a,b), (c,d), (e,f) is given by
$x=\frac{a+c+e}{3}$
$y=\frac{b+d+f}{3}$
In the given question:
$1=\frac{h+2-6}{3}$
Solving for $h$ we get
$h =7 $
$5=\frac{-6+3+ k }{3}$
Solving for k we get
$k=18$
View full question & answer→Question 62 Marks
In ΔABC, G (-4, -7) is the centroid. If A (-14, -19) and B(3, 5) then find theco-ordinates of C.
AnswerThe coordinates of the centroid ( $x, y$ ) od a triangle formed by points $(a, b),(c, d),(e, f)$ is given by
$x=\frac{a+c+e}{3} $
$y=\frac{b+d+f}{3}$
In the given question $(x, y)=(-4,-7)$
Hence $-4=\frac{-14+3+e}{3}$
Solving for e, we get
$e=-1$
$-7=\frac{-19+5+f}{3}$
Solving for f , we get
$f=-7$
Hence the coordinates of the third point are $(-1,-7)$
View full question & answer→Question 72 Marks
Find the coordinates of midpoint of the segment joining the points (22,20) and (0, 16).
AnswerAccording to the mid point theorem the coordinates of the point P(x,y) dividing the line formed by A(a,b) and B(c,d) is given by:
$x =\frac{ a + c }{2} $
$y =\frac{ b + d }{2}$
The coordinates of midpoint $(x, y)$ are
$x=\frac{22+0}{2}=11 $
y=\frac{20+16}{2}=18$
Hence the coordinates are (11,18)
View full question & answer→Question 82 Marks
Find the ratio in which point P(k, 7) divides the segment joining A(8, 9) andB(1, 2). Also find k.
AnswerA point P(x,y) divides the line formed by points (a,b) and (c,d) in the ratio of m:n, then the coordinates of the point P is given by
$x =\frac{ an + cm }{ m + n } \text { and } y =\frac{ bn + dm }{ m + n }$
In the given question,
Let the point P divide AB is the ratio 1:k
Y coordinate of P
$7=\frac{9 k+2}{k+1}$
Simplifying
7k + 7 = 9k + 2
2k = 5
$7 k +7=9 k +2$
$2 k =5$
$k =\frac{5}{2}$
Therefore point P divides AB in the ratio 2:5
View full question & answer→Question 92 Marks
Find the slope of the diagonals of a quadrilateral with verticesA(1,7), B(6,3), C(0,-3) and D(-3,3).
AnswerSlope of a line between two points $(a, b)$ and $(c, d)$ is
$m =\frac{ d - b }{ c - a }$
A quadrilateral $A B C D$ has diagonals $A C$ and $B D$
Slope $A C=\frac{-3-7}{0-1}=10$
Slope $B D=\frac{3-3}{-3-6}=0$
View full question & answer→Question 102 Marks
Point P is the centre of the circle and AB is a diameter. Find the coordinates ofpoint B if coordinates of point A and P are (2, -3) and (-2, 0) respectively.
AnswerAccording to the mid point theorem the coordinates of the point P(x,y) dividing the line formed by A(a,b) and B(c,d) is given by:
$x=\frac{a+c}{2}$
$y=\frac{b+d}{2}$
In the given question A = (1,-3) and midpoint P is (-2,0).
Let coordinates of B be (c,d)
Then,
$-2=\frac{2+c}{2}$
And
$0=\frac{-3+d}{2}$
Solving for c and d, we get
-4 = 2 + c
c = -6
d = 3
Hence the coordinates of point B are (-6,3).
View full question & answer→Question 112 Marks
Given A(4,-3), B(8,5). Find the coordinates of the point that divides segmentAB in the ratio 3 : 1.
AnswerA point P(x,y) divides the line formed by points (a,b) and (c,d) in the ratio of m:n, then the coordinates of the point P is given by
$X=\frac{a n+c m}{m+n} \text { and } y=\frac{b n+d m}{m+n} $
$X \text { coordinate }=\frac{(4 \times 1+(8 \times 3))}{(3+1)}=7 $
$ Y \text { coordinate }=\frac{(-3) 1+(5) 3}{(1+3)}=3$
Hence the point is (7,3)
View full question & answer→Question 122 Marks
Show that the points A(1, 2), B(1, 6), C(1 + 2 √3, 4) are vertices of an equilateral triangle.
AnswerFor an equilateral triangle, all its sides are equal.
According to the distance formula, the distance ' $d$ ' between two points $(a, b)$ and $(c, d)$ is given by
$d=\sqrt[2]{(a-c)^2+(b-d)^2} \ldots$
Length $A B=\sqrt{(1-1)^2+(6-2)^2}=\sqrt{16}=4$
Length $B C=\sqrt{(1+2 \sqrt{3}-1)^2+(4-6)^2}=\sqrt{12+4}=4$
Length $A C=\sqrt{(1+2 \sqrt{3}-1)^2+(4-2)^2}=\sqrt{12+4}=4$
Hence The given points form an equilateral triangle.
View full question & answer→Question 132 Marks
Find x if distance between points L(x, 7) and M(1, 15) is 10.
AnswerAccording to the distance formula, the distance 'd' between two points (a,b) and (c,d) is given by
$d=\sqrt[2]{(a-c)^2+(b-d)^2} \ldots(1)$
Distance between $LM =\sqrt{( x -1)^2+(7-15)^2}=10$
Squaring both sides, we get
$(x-1)^2+64=100$
$(x-1)^2=36$
$x-1= \pm 6$
Hence $x=7$ or -5
View full question & answer→Question 142 Marks
Find the coordinates of centroid of the triangles if points D(-7,6),E(8,5) andF(2, -2) are the mid points of the sides of that triangle.
AnswerThe coordinates of the centroid (x,y) od a triangle formed by points (a,b), (c,d), (e,f) is given by
$x=\frac{a+c+e}{3} $
$ y=\frac{b+d+f}{3} $
$X \text { coordinate }=\frac{-7+8+2}{3}=1 $
$ Y \text { coordinate }=\frac{6+5-2}{3}=3$
Hence coordinates are (1,3)
View full question & answer→Question 152 Marks
Show that the line joining the points A(4, 8) and B(5, 5) is parallel to the linejoining the points C(2,4) and D(1,7).
AnswerSlope of a line between two points (a,b) and (c,d) is
$m =\frac{ d - b }{ c - a }$
Slope of $A B=\frac{5-8}{5-4}=-3$
Slope of $A C=\frac{7-4}{1-2}=-3$
As slopes are equal, the two lines are parallel.
View full question & answer→Question 162 Marks
Find k if the line passing through points P(-12,-3) and Q(4, k)has slope $\frac{1}{2}$.
AnswerSlope of a line between two points (a,b) and (c,d) is
$ m =\frac{ d - b }{ c - a } $
$ \text { Slope }=\frac{ k -(-3)}{4-(-12)}=\frac{1}{2}$
Simplifying
K = 5
View full question & answer→Question 172 Marks
If $A(6,1), B(8,2), C(9,4)$ and $D(7,3)$ are the vertices of $\square \mathrm{ABCD}$, show that $\square$ ABCD is a parallelogram.
AnswerYou know that Slope of line $=\frac{y_2-y_1}{x_2-x_1}$
Slope of line $A B=\frac{2-1}{8-6}=\frac{1}{2}$
Slope of line $\mathrm{BC}=\frac{4-2}{9-8}=2$
Slope of line $C D=\frac{3-4}{7-9}=\frac{1}{2}$
Slope of line DA $=\frac{3-1}{7-6}=2$
Slope of line $\mathrm{AB}=$ Slope of line $\mathrm{CD}$ From (I) and (III)
$\therefore$ line $\mathrm{AB} \|$ line $\mathrm{CD}$
Slope of line $\mathrm{BC}=$ Slope of line DA From (II) and (IV)
$\therefore$ line $\mathrm{BC} \|$ line DA
Both the pairs of opposite sides of the quadrilateral are parallel $\therefore \square \mathrm{ABCD}$ is a parallelogram.
View full question & answer→Question 182 Marks
$A(15,5), B(9,20)$ and $A-P-B.$ Find the ratio in which point $P(11,15)$ divides segment $AB.$
AnswerSuppose, point $\mathrm{P}(11,15)$ divides segment $\mathrm{AB}$ in the ratio $m: n$
$\therefore$ by section formula,
$ x=\frac{m x_2+n x_1}{m+n}$
$\therefore 11=\frac{9 m+15 n}{m+n}$
$\therefore 11 m+11 n=9 m+15 n$
$\therefore 2 m=4 n$
$\therefore \frac{m}{n}=\frac{4}{2}=\frac{2}{1} $
Similarly, find the ratio using $y$ co-ordinates. Write the conclusion.
$\therefore$ The required ratio is $2: 1$.
View full question & answer→Question 192 Marks
Find the co-ordinates of point $\mathrm{P}$ if $\mathrm{P}$ is the midpoint of a line segment $\mathrm{AB}$ with $A(-4,2)$ and $B(6,2)$.
AnswerIn the given example, suppose
Fig. 5.15
$
\begin{array}{c}
(-4,2)=\left(x_1, y_1\right) ;(6,2)=\left(x_2, y_2\right) \text { and co-ordinates of } \mathrm{P} \text { are }(x, y) \\
\therefore \text { according to midpoint theorem, } \\
x=\frac{x_1+x_2}{2}=\frac{-4+6}{2}=\frac{2}{2}=1 \\
y=\frac{y_1+y_2}{2}=\frac{2+2}{2}=\frac{4}{2}=2 \\
\therefore \text { co-ordinates of midpoint } \mathrm{P} \text { are }(1,2) .
\end{array}
$
View full question & answer→Question 202 Marks
If $A(3,5), B(7,9)$ and point $Q$ divides seg $A B$ in the ratio $2: 3$ then find co-ordinates of point $Q$.
AnswerIn the given example let $\left(x_1, y_1\right)=(3,5)$
$\text { and }\left(x_2, y_2\right)=(7,9) . $
$m: n=2: 3$
According to section formula,
$x=\frac{m x_2+n x_1}{m+n}=\frac{2 \times 7+3 \times 3}{2+3}=\frac{23}{5} \quad y=\frac{m y_2+n y_1}{m+n}=\frac{2 \times 9+3 \times 5}{2+3}=\frac{33}{5}$
$\therefore \text { Co-ordinates of } \mathrm{Q} \text { are }\left(\frac{23}{5}, \frac{33}{5}\right) $
View full question & answer→Question 212 Marks
Find the value of $y$ if distance between points $\mathrm{A}(2,-2)$ and $\mathrm{B}(-1, y)$ is $5 .$
Answer$ \mathrm{AB}^2=[(-1)-2]^2+[y-(-2)]^2 .$
$\therefore \quad 5^2=(-3)^2+(y+2)^2$
$\therefore 25=9+(y+2)^2$
$\therefore 16=(y+2)^2$
$\therefore y+2= \pm \sqrt{16}$
$\therefore y+2= \pm 4$
$\therefore y=4-2 \text { or } y=-4-2$
$\therefore y=2 \text { or } y=-6$
$\therefore \text { value of } y \text { is } 2 \text { or }-6 .$
by distance formula
View full question & answer→Question 222 Marks
If point $(x, y)$ is equidistant from points $(7,1)$ and $(3,5)$, show that $y=x-2$.
AnswerLet point $\mathrm{P}(x, y)$ be equidistant from points $\mathrm{A}(7,1)$ and $\mathrm{B}(3,5)$
$
\begin{aligned}
\therefore \mathrm{AP} & =\mathrm{BP} \\
\therefore \mathrm{AP}^2 & =\mathrm{BP}^2 \\
\therefore(x-7)^2+(y-1)^2 & =(x-3)^2+(y-5)^2 \\
\therefore x^2-14 x+49+y^2-2 y+1 & =x^2-6 x+9+y^2-10 y+25 \\
\therefore-8 x+8 y & =-16 \\
\therefore x-y & =2 \\
\therefore y & =x-2
\end{aligned}
$
View full question & answer→Question 232 Marks
Find the co-ordinates of a point on $Y$ - axis which is equidistant from $\mathrm{M}(-5,-2)$ and $\mathrm{N}(3,2)$
AnswerLet point $\mathrm{P}(0, y)$ on $\mathrm{Y}$ - axis be equidistant from $\mathrm{M}(-5,-2)$ and $\mathrm{N}(3,2)$.
$
\begin{array}{c}
\therefore \mathrm{PM}=\mathrm{PN} \quad \therefore \mathrm{PM}^2=\mathrm{PN}^2 \\
\therefore[0-(-5)]^2+[y-(-2)]^2=(0-3)^2+(y-2)^2 \\
\therefore 25+(y+2)^2=9+y^2-4 y+4 \\
\therefore 25+\mathrm{y}^2+4 y+4=13+y^2-4 y \\
\therefore 8 y=-16 \quad \therefore y=-2
\end{array}
$
$\therefore$ the co-ordinates of the point on the $\mathrm{Y}$-axis which is equidistant from $\mathrm{M}$ and $\mathrm{N}$ are $\mathrm{M}(0,-2)$.
View full question & answer→Question 242 Marks
Find the distance between the points $P(-1,1)$ and $Q(5,-7)$.
AnswerSuppose co-ordinates of point $\mathrm{P}$ are $\left(x_1, y_1\right)$ and of point $\mathrm{Q}$ are $\left(x_2, y_2\right)$.
$
x_1=-1, \quad y_1=1, \quad x_2=5, \quad y_2=-7
$
According to distance formula, $\mathrm{d}(\mathrm{P}, \mathrm{Q})=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$
\begin{aligned}
& =\sqrt{[5-(-1)]^2+[(-7)-1]^2} \\
& =\sqrt{(6)^2+(-8)^2} \\
& =\sqrt{36+64} \\
d(P, Q) & =\sqrt{100}=10
\end{aligned}
$
$\therefore$ distance between points $\mathrm{P}$ and $\mathrm{Q}$ is 10 .
View full question & answer→Question 252 Marks
Find the value of $k$, if the slope of the line passing through (2, 5) and (k, 3) is 2.
View full question & answer→Question 262 Marks
Using slope concept, check whether the following points are collinear.
(i) $(-2,-1)(4,0)(3,3)$
(ii) $(-2,-3),\left(\frac{ 3 3 }{ 8 }, 4\right)(5,5)$
(iii) $(4,4)(3,5)(-1,-1)$
(iv) $(2,10),(0,4)(3,13)$
(v) $(5,0)(10,-3)(-5,6)$
(vi) $(2,5),(5,7)(8,9)$
Answer(ii), (iv), (v), (vi) are collinear.
View full question & answer→Question 272 Marks
Find the slope of the line passing through the points.
(i) $(-1,4)(3,-7)$
(ii) $(5,5),(1,6)$
(iii) $(1,7)(4,8)$
(iv) $(4,8),(5,5)$
(v) $(4,1)(2,-3)$
(vi) $(4,4),(3,5)$
Answer(i) $-1 \frac{1}{4}$, (ii) $-\frac{1}{4}$, (iii) $\frac{1}{3}$, (iv) -3, (v) 2, (vi) -1
View full question & answer→Question 282 Marks
Find the coordinates of the midpoint P of seg AB , if $A (3.5,9.5)$ and $B (-1.5,0.5) \quad$
View full question & answer→Question 292 Marks
Find the coordinates of the midpoint of segment $Q R$, if $Q(2.5,-4.3)$ and $R(-1.5,2.7)$
View full question & answer→Question 302 Marks
Find the coordinates of the point P which divides line segment QR in the ratio m : n in the following examples.
(i) $Q (-5,8), \quad R (4,-4) \quad m: n=2: 1$
(ii) $Q (-2,7), \quad R (-2,-5) \quad m: n=1: 3$
(iii) $Q (1,7), \quad R (-3,1) \quad m: n=1: 2$
(iv) $Q (6,-5), \quad R (-10,2) \quad m: n=3: 4$
(v) $Q (5,8), \quad R (-7,-8) \quad m: n=4: 1$
Answer(i) $(1,0)$, (ii) $(-2,4)$, (iii) $\left(-\frac{1}{3}, 5\right)$, (iv) $\left(-\frac{6}{7},-2\right)$, (v) $\left(\frac{23}{5},-\frac{24}{5}\right)$
View full question & answer→Question 312 Marks
Find the distance of point Z(-2.4, -1), from the origin.
Question 322 Marks
Using distance formula, check whether following points are collinear or not.
(i) L(4,-1) M(1, -3), N(-2,-5)
(ii) A(-5, 4), B(-2, -2), C(3, -12)
Answer(i) collinear, (ii) non-collinear
View full question & answer→Question 332 Marks
Show that the point (0, 9) is equidistant from the point (-4, 1) and (4, 1)
Question 342 Marks
Check whether points (3, 3), (-4, -1) and (3, -5) are the vertices of an isosceles triangle.
Question 352 Marks
Show that the point (5, 11) is equidistant from the points (-5, 13) and (3, 1).
Question 362 Marks
Find the ratio in which the line segment joining the points A(3,8) and B(-9, 3)is divided by the Y-axis.
Question 372 Marks
Find k, if PQ || RS and P(2, 4), Q (3, 6), R(3, 1), S(5, k) .
Question 382 Marks
Find k, if B(k, -5), C (1, 2) and slope of the line is 7.
Question 392 Marks
Find k, if R(1, -1), S (-2, k) and slope of line RS is -2.
Question 402 Marks
A(h, -6), B(2, 3) and C(-6, k) are the co-ordinates of vertices of a trianglewhose centroid is G (1, 5). Find h and k.
Question 412 Marks
In ΔABC, G (-4, -7) is the centroid. If A (-14, -19) and B(3, 5) then find theco-ordinates of C.
Question 422 Marks
Find the coordinates of midpoint of the segment joining the points (22,20) and (0, 16).
Question 432 Marks
Find the ratio in which point P(k, 7) divides the segment joining A(8, 9) andB(1, 2). Also find k.
Question 442 Marks
Find the slope of the diagonals of a quadrilateral with verticesA(1,7), B(6,3), C(0,-3) and D(-3,3).
Question 452 Marks
Point P is the centre of the circle and AB is a diameter. Find the coordinates ofpoint B if coordinates of point A and P are (2, -3) and (-2, 0) respectively.
Question 462 Marks
Given A(4,-3), B(8,5). Find the coordinates of the point that divides segmentAB in the ratio 3 : 1.
Question 472 Marks
Show that the points A(1, 2), B(1, 6), C(1 + 2 √3, 4) are vertices of an equilateral triangle.
Question 482 Marks
Find x if distance between points L(x, 7) and M(1, 15) is 10.
Question 492 Marks
Find the coordinates of centroid of the triangles if points D(-7,6),E(8,5) andF(2, -2) are the mid points of the sides of that triangle.
Question 502 Marks
Show that the line joining the points A(4, 8) and B(5, 5) is parallel to the linejoining the points C(2,4) and D(1,7).