Questions & Answers

Question

Answers

A.$\left( {6,4} \right)$

B.$\left( { - 3,3} \right)$

C.$\left( { - 8,8} \right)$

D.$\left( {0,7} \right)$

Answer
Verified

The general equation of the family of lines through the point of intersection of two given lines L1 and L2 is given by

${L_1} + \lambda {L_2} = 0$, where $\lambda $ is a parameter.

Equation of a line passing through two points $\left( {{x_1},{y_1}} \right)\& \left( {{x_2},{y_2}} \right)$ is given as

$y - {y_1} = \left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)\left( {x - {x_1}} \right)$

Two lines L1 and L2 are perpendicular to each other if ${m_1}{m_2} = - 1$

According to the question,

Lines $2x + 3y - 1 = 0,x + 2y - 1 = 0$ and $ax + by - 1 = 0$ form a triangle, so to find the intersection point of lines $2x + 3y - 1 = 0,x + 2y - 1 = 0$ we have to convert the given equation into slope intercept form i.e. $y = mx + c$

Now, the equations are as follows

$\begin{gathered}

3y = 1 - 2x \ldots \left( 1 \right) \\

2y = 1 - x \\

x = 1 - 2y \ldots \left( 2 \right) \\

\end{gathered} $

Substitute the value of x in equation (1)

$\begin{gathered}

3y = 1 - 2\left( {1 - 2y} \right) \\

3y = 1 - 2 + 4y \\

y = 1 \\

\end{gathered} $

Put the value of y in equation (2)

$\begin{gathered}

x = 1 - 2y \\

x = 1 - 2\left( 1 \right) \\

x = - 1 \\

\end{gathered} $

Hence, the intersection point is (-1, 1)

Line passing through (-1, 1) and origin is perpendicular to $ax + by - 1 = 0$

Required line ${L_1}:y = - x$ and ${L_2}:ax + by - 1 = 0$ are perpendicular as the product of their slope is -1.

By using ${m_1}{m_2} = - 1$, we get $a = - b \ldots \left( i \right)$

Also line passing through $x + 2y - 1 = 0$ and $ax + by - 1 = 0$, is

$x + 2y - 1 + \lambda \left( {ax + by - 1} \right) = 0$

It passes through origin, so

$\lambda = - 1$

Line is $x\left( {1 - a} \right) + y\left( {2 - b} \right) = 0$

This is perpendicular to line $2x + 3y - 1 = 0$

So, using ${m_1}{m_2} = - 1$, we get

$2a + 3b = 8 \ldots \left( {ii} \right)$

Then on solving equation (i) and (ii), we get

$\begin{gathered}

a = 8 \\

b = - 8 \\

\end{gathered} $

Hence, (a, b) is (8, -8)

1)For an acute triangle, it lies inside the triangle.

2)For an obtuse triangle, it lies outside of the triangle.

3)For a right-angled triangle, it lies on the vertex of right angle