The ends of the base of an isosceles triangle are at $(2 a, 0)$ and $(0, a)$ . The equation of one side is $x = 2 a$ . The equation of the other side is

x + 2 y - a = 0

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x + 2 y = 2 a

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3 x + 4 y - 4 a = 0

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3 x - 4 y + 4 a = 0

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Solution

The correct option is D $3 x - 4 y + 4 a = 0$
Let third vertex be $(2 a, h)$
Now $P Q = P R$
$∴ 4 a^{2} + (h - a)^{2} = h^{2}$
$h = \frac{5 a}{2}$
$∴$ Third line $P Q$ is
$3 x - 4 y + 4 a = 0$

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Similar questions

Q. If the extremities of the base of an isosceles triangle are the points

(2 a, 0)

and

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and the equation of one of the sides is

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, then the area of the triangle is

Q. If the extremities of the base of an isosceles triangle are the points

(2 a, 0)

and

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and the equation of one of the sides is

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and parallel to the straight line

3 x + 4 y = 0

If the area of the parallelogram whose sides are x + 2y + 3 = 0, 3x + 4y - 5 = 0, 2x+4y+5=0 and 3x + 4y - 10 = 0 is 'a' sq. unit. Find the value of 4a

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Q. Find the area of the parallelogram whose sides are:

x + 2 y + 3 = 0, 3 x + 4 y - 5 = 0, 2 x + 4 y + 5 = 0 a n d 3 x + 4 y - 10 = 0

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The ends of the base of an isosceles triangle are at (2a,0) and (0,a). The equation of one side is x=2a. The equation of the other side is

The correct option is D 3x−4y+4a=0Let third vertex be (2a,h)Now PQ=PR∴4a2+(h−a)2=h2h=5a2∴ Third line PQ is3x−4y+4a=0

The ends of the base of an isosceles triangle are at $(2 a, 0)$ and $(0, a)$ . The equation of one side is $x = 2 a$ . The equation of the other side is

The correct option is D $3 x - 4 y + 4 a = 0$
Let third vertex be $(2 a, h)$
Now $P Q = P R$
$∴ 4 a^{2} + (h - a)^{2} = h^{2}$
$h = \frac{5 a}{2}$
$∴$ Third line $P Q$ is
$3 x - 4 y + 4 a = 0$