[4 marks sum]

Question 14 Marks

The points $A (3, 0), B (a, -2)$ and $C (4, -1)$ are the vertices of $\triangle ABC$ right angled at vertex $A$. Find the value of $a.$

Answer

$A B=\sqrt{(a-3)^2+(-2-0)^2}$
$=\sqrt{a^2+9-6 a+4}$
$=\sqrt{a^2-6 a+13}$
$B C=\sqrt{(4-a)^2+(-1+2)^2}$
$=\sqrt{a^2+16-8 a+1}$
$=\sqrt{a^2-8 a+17}$
$ CA =\sqrt{(3-4)^2+(0+1)^2}$
$=\sqrt{1+1}$
$=\sqrt{2}$
Since, $\triangle ABC$ is a right$-$angled at $A,$ we have:
$AB^2 + AC^2 = BC^2$
$\Rightarrow a^2 - 6a + 13 + 2 = a^2 - 8a + 17$
$\Rightarrow 2a = 2$
$\Rightarrow a = 1$

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Question 24 Marks

The vertices of a triangle are $(5, 1), (11, 1)$ and $(11, 9)$. Find the co$-$ordinates of the circumcentre of the triangle.

Answer

Let the circumcentre be $\mathrm{P}(\mathrm{x}, \mathrm{y})$.
Then, $\mathrm{PA}=\mathrm{PB}$
$\mathrm{PA}^2=\mathrm{PB}^2$
$(x-5)^2+(y-1)^2=(x-11)^2+(y-1)^2$
$x^2+25-10 x=x^2+121-22 x$
$12 x=96$
$x=8$
Also, $\mathrm{PA}=\mathrm{PC}$
$P A^2=P C^2$
$(x-5)^2+(y-1)^2=(x-11)^2+(y-9)^2$
$x^2+25-10 x+y^2+1-2 y=x^2+121-22 x+y^2+81-18 y$
$12 x+16 y=176$
$3 x+4 y=44$
$24+4 y=44$
$4 y=20$
$y=5$
Thus, the coordinates of the circumcentre of the triangle are $(8,5)$.

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Question 34 Marks

Prove that the points $A (1, -3), B (-3, 0)$ and $C (4, 1)$ are the vertices of an isosceles right$-$angled triangle. Find the area of the triangle.

Answer

$ AB =\sqrt{(-3-1)^2+(0+3)^2}=\sqrt{16+9}=\sqrt{25}=5$
$BC =\sqrt{(4+3)^2+(1+0)^2}=\sqrt{49+1}=\sqrt{50}=5 \sqrt{2}$
$CA =\sqrt{(1-4)^2+(-3-1)^2}=\sqrt{9+16}=\sqrt{25}=5$
$\because AB = CA$
$A, B, C$ are the vertices of an isosceless triangle.
$AB^2 + CA^2 = 25 + 25 = 50$
$ BC ^2=(5 \sqrt{2})^2=50$
$\therefore AB ^2+ CA ^2= BC ^2$
Hence,$ A, B., C$ are the vertices of a right $-$ angled triangle.
Hence, $\triangle ABC$ is an isosceles right$-$angled triangle.
Area of $\triangle ABC =\frac{1}{2} \times AB \times CA$
$=\frac{1}{2} \times 5 \times 5$
$=12.5 \text { sq}$.units

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Question 44 Marks

A point $P (2, -1)$ is equidistant from the points $(a, 7)$ and $(-3, a)$. Find $a.$

Answer

We know that the distance between the two points $(x1, y1)$ and $(x2, y2)$ is
$d =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
Let the given points be $A = (a, 7)$ and $B = (−3, a)$ and the third point given is $P(2, −1).$
We first find the distance between $P(2, −1)$ and $A =(a, 7)$ as follows:
$ PA =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$=\sqrt{(a-2)^2+(7-(-1))^2}$
$=\sqrt{(a-2)^2+(7+1)^2}$
$=\sqrt{(a-2)^2+8^2}$
$=\sqrt{(a-2)^2+64}$
Similarly, the distance between $P(2,−1)$ and $B = (−3, a)$ is:
$ PB =\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$=\sqrt{(-3-2)^2+(a-(-1))^2}$
$=\sqrt{(-5)^2+(a+1)^2}$
$=\sqrt{25+(a+1)^2}$
Since the point $P(2,−1)$ is equidistant from the points $A(a, 7)$ and $B = (−3, a),$
$\therefore, PA = PB$ that is :
$\rightarrow \sqrt{(a-2)^2+64}=\sqrt{25+(a+1)^2}$
$\rightarrow \left(\sqrt{(a-2)^2+64^2}\right)=\left(\sqrt{25+\left((a+1)^2\right)^2}\right.$
$\Rightarrow (a − 2)^2+ 64 = 25 + (a + 1)^2$
$\Rightarrow (a − 2)^2− (a + 1)^2= 25 − 64$
$\Rightarrow (a^2+ 4 − 4a) − (a^2+ 1 + 2a) = −39 \dots...( \because (a − b)^2= a^2+ b^2− 2ab, (a + b)^2= a^2+ b^2+ 2ab)$
$\Rightarrow a^2+ 4 − 4a − a^2− 1 − 2a = −39$
$\Rightarrow −6a + 3 = −39$
$\Rightarrow − 6a =−39 −3$
$\Rightarrow −6a = −42$
$\Rightarrow a =\frac{-42}{-6}=7$
Hence$, a = 7.$

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Question 54 Marks

A point $A$ is at a distance of $\sqrt{10}$ unit from the point $(4, 3).$ Find the co$-$ordinates of point $A,$ if its ordinate is twice its abscissa.

Answer

It is given that the coordinates of point $A$ are such that its ordinate is twice its abscissa.
So, let the coordinates of point $A$ be $(x, 2x).$
We have :
$\sqrt{(x-4)^2+(2 x-3)^2}=\sqrt{10}$
$(x - 4)^2 + (2x - 3)^2= 10$
$x^2 + 16 - 8x + 4x^2 + 9 - 12x = 10$
$5x^2 - 20x + 25 = 10$
$5x^2 - 20x = 10 - 25$
$5x^2 - 20x = - 15$
$5x^2 - 20x + 15 = 0$
$x^2 - 4x + 3 = 0$
$x^2- x - 3x + 3 = 0$
$x(x - 1) -3(x - 1) = 0$
$(x - 1)(x - 3) = 0$
$x = 1, 3$
Thus, the co$-$ordinates of the point $A$ are $(1, 2)$ and $(3, 6).$

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Question 64 Marks

Find the point on $y-$axis whose distances from the points $A (6, 7)$ and $B (4, -3)$ are in the ratio $1: 2.$

Answer

Let the required point on $y-$axis be $P (0, y).$
$ PA =\sqrt{(0-6)^2+(y-7)^2}$
$=\sqrt{36+y^2+49-14 y}$
$=\sqrt{y^2-14 y+85}$
$PB =\sqrt{(0-4)^2+(y+3)^2}$
$=\sqrt{16+y^2+9+6 y}$
$=\sqrt{y^2+6 y+25}$
From the given information, we have:
$\frac{ PA }{ PB }=\frac{1}{2}$
$\frac{ PA ^2}{ PB ^2}=\frac{1}{4}$
$\frac{y^2-14 y+85}{y^2+6 y+25}=\frac{1}{4}$
$4y^2 - 56y + 340$
$= y^2+ 6y + 25$
$3y^2 - 62y + 315$
$= 0$
$y=\frac{62 \pm \sqrt{3844-3780}}{6}$
$y=\frac{62 \pm 8}{6}$
$y=9, \frac{35}{3}$
Thus, the required points on $y-$axis are $(0, 9)$ and $\left(0, \frac{35}{3}\right)$.

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MATHEMATICS [4 marks sum]

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