The Straight Lines — MATHS STD 11 Science — Question
Gujarat BoardEnglish MediumSTD 11 ScienceMATHSThe Straight Lines4 Marks
Question
Show that the perpendicular bisectors of the sides of a triangle are concurrent.
✓
Answer
Let coordinates of $\triangle\text{ABC}$ A(0, 0), B(a, 0), c(0, b). Then mid points of AB, BC and CA are → $\text{D}\Big(\frac{\text{a}}{2},0\Big), \text{E}\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)$ and $\text{F}\Big(0,\frac{\text{b}}{2}\Big)$ Then equation of CD, AE and BF are $\text{CD}\Rightarrow\text{y}-\text{b}=\frac{\text{a}-\text{b}}{\frac{\text{a}}{2}-0}(\text{x}-0)$ $\Rightarrow\text{y}-\text{b}=\frac{-2\text{b}}{\text{a}}(\text{x})$ $\Rightarrow\text{ay}-\text{ab}=-2\text{bx}$ $\Rightarrow\text{ay}+2\text{bx}-\text{ab}=0 \ ...(1)$ $\text{BF}\Rightarrow\text{y}-0=\frac{\frac{\text{b}}{2}-0}{0-\text{a}}(\text{x}-0)$ $\Rightarrow\text{y}=\frac{-\text{b}}{2\text{a}}(\text{x}-\text{a})$ $\Rightarrow-2\text{ay}-\text{bx}=\text{ba} \ ...(2)$ $\text{AE}\Rightarrow\text{y}-0=\frac{0-\frac{\text{b}}{2}}{0-\frac{\text{a}}{2}}(\text{x}-0)$ $\Rightarrow\text{ya}=+\text{bx} \ ...(3)$ Adding (1), (2) and (3) $\text{ay}+2\text{bx}-\text{ab}+2\text{b}^2-2\text{ay}-\text{bx}-\text{ab}+\text{ay}-\text{bx}=0$ then, $\lambda_1\text{L}_1+\lambda_2\text{L}_2+\lambda_3\text{L}_3=0.$ where $\lambda_1=\lambda_2=\lambda_3=1.$ Hence, lines are concurrent.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.