Question

The straight lines given by the equations $x+y=2,x-2y=5and\frac{x}{3}+y=0$ are?

• A. concurrent
• B. intersecting to make a right triangle.
• C. intersecting to make an isosceles triangle.
• D. parallel to each other.

Answer 1

Answer: (A)

concurrent

Given lines

$x+y=2$------(1)

$x-2y=5$----(2) and $\frac{x}{3}+y=0$----(3)

Solving eq (1) and (2)

$x-2(2-x)=5$

$x-4+2x=5$

$x=3$ and $y=2-3=-1$

Point of intersection of line (1) and (2) is $P(3,-1)$

Solving eq (2) and (3)

$-3y-2y=5$

$-5y=5$

$y=-1$ and $x=-3y=3$

Point of intersection of line (2) and (3) is $Q(3,-1)$

Solving eq (1) and (3)

$-3y+y=2$

$-2y=2$

$y=-1$ and $x=-3y=3$

Point of intersection of line (1) and (3) is $R(3,-1)$

Here point of intersection of all line is same Hence line is concurrent