First slide

Equation of line in Straight lines

Question

The equation of the sides of an Isosceles right angled triangle whose hypotenuse is $7 x + y - 8 = 0$ and opposite vertex to the hypotenuse is $(1, 1)$ are

Moderate

A

$3 x + 4 y - 7 = 0, 4 x - 3 y - 1 = 0$
B

$3 x - 4 y - 7 = 0, 4 x + 3 y - 1 = 0$
C

$3 x - 4 y + 7 = 0, 4 x + 3 y = 1$
D

$3 x - 4 y - 8 = 0, 4 x + 3 y - 2 = 0$

Solution

$The equation of any line which makes an angle α with the line a x + b y + c = 0 and passing$
$through the point (x_{1}, y_{1}) can be taken as y - y_{1} = \tan (θ \pm α) (x - x_{1}), here θ is the$
$inclination of the line a x + b y + c = 0$
$The slope of the line a x + b y + c = 0 is m = - \frac{a}{b} = \tan θ$
$Since the hypotenuse of right angled isosceles triangle is 7 x + y - 8 = 0 and opposite vertex is$
$(1,1) then the other two sides makes equal angles 45^{\circ} each with the hypotenuse and passing$
$through the point (1, 1)$
$Hence, \tan θ = - 7, α = 45^{\circ}$
$Therefore, the equation of one side is$
$\begin{matrix} y - 1 = \tan (θ + 45 °) (x - 1) \\ = \frac{1 + \tan θ}{1 - \tan θ} (x - 1) \\ = - \frac{3}{4} (x - 1) \end{matrix}$
It implies,
$\begin{matrix} 4 y - 4 = - 3 x + 3 \\ 3 x + 4 y - 7 = 0 \end{matrix}$
And the equation of the other side is
$\begin{matrix} y - 1 = \tan (θ - 45 °) (x - 1) \\ = \frac{\tan θ - 1}{\tan θ + 1} (x - 1) \\ = \frac{4}{3} (x - 1) \end{matrix}$
It implies,
$\begin{matrix} 3 y - 3 = 4 x - 4 \\ 4 x - 3 y - 1 = 0 \end{matrix}$
$Therefore, two sides of given triangle are 3 x + 4 y - 7 = 0, 4 x - 3 y - 1 = 0$

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