Mathematics - Straight Lines Question with Solution | TestHub

Let the coordinates of $A$ be $(x_A,y_A)$ and $B$ be $(x_B,y_B)$. Since $P(1,1)$ is the midpoint of $AB$:$$\frac{x_A+x_B}{2}=1⟹x_A+x_B=2$$ $$\frac{y_A+y_B}{2}=1⟹y_A+y_B=2$$ Point$A(x_A,y_A)$ lies on $L_1:x_A+2y_A-3=0$. Point $B(x_B,y_B)$ lies on $L_2:2x_B-y_B+1=0$. From the midpoint equations, $x_A=2-x_B$ and $y_A=2-y_B$. Substitute these into the equation for $A$:$(2-x_B)+2(2-y_B)-3=0⟹2-x_B+4-2y_B-3=0⟹3-x_B-2y_B=0⟹x_B+2y_B-3=0$.

Now we have a system of two equations for $x_B$ and $y_B$: 1)$x_B+2y_B-3=0$ 2)$2x_B-y_B+1=0$ From (2),$y_B=2x_B+1$. Substitute into (1): $x_B+2(2x_B+1)-3=0⟹x_B+4x_B+2-3=0⟹5x_B-1=0⟹x_B=1/5$. Then $y_B=2(1/5)+1=7/5$. So, point $B$ is $(1/5,7/5)$.

The line $L$ passes through $P(1,1)$ and $B(1/5,7/5)$. The slope of $L$ is $m=\frac{7/5-1}{1/5-1}=\frac{2/5}{-4/5}=-1/2$. The equation of line $L$ is $y-1=-1/2(x-1)⟹2(y-1)=-(x-1)⟹2y-2=-x+1⟹x+2y-3=0$.