Equations of two equal sides of an isosceles triangle are 7x - y + 3 = 0 and x + y - 3 = 0 and third

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9.Straight Line

hard

The equations of two equal sides of an isosceles triangle are $7x - y + 3 = 0$ and $x + y - 3 = 0$ and the third side passes through the point $(1, -10)$. The equation of the third side is

A

$y = \sqrt 3 x + 9$ but not $y = -\sqrt 3 x + 9$

B

$3x + y + 7 = 0$ but not $3x + y - 7 = 0$

C

$3x + y + 7 = 0$ or $x - 3y - 31 = 0$

D

Neither $3x + y + 7$ nor $x - 3y - 31 = 0$

(IIT-1984)

Solution

(c) Any line through $(1, -10)$ is given by $y + 10 = m(x – 1)$

Since it makes equal angle say ‘$\alpha $’ with the given lines $7x – y + 3 = 0$ and $x + y – 3 = 0$, therefore

$\tan \alpha = \frac{{m – 7}}{{1 + 7m}}$ $ = \frac{{m – ( – 1)}}{{1 + m( – 1)}} \Rightarrow m = \frac{1}{3}$ or $-3$

Hence the two possible equations of third side are $3x + y + 7 = 0$ and $x – 3y – 31 = 0$.

Standard 11

Mathematics

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