Abhi koi question add nahi kiya. PYQ Bank se questions add karein.
PYQ Bank ›
Class 10 › Mathematics › Coordinate Geometry
CBSE2026Class Class 10 · Mathematics3 Marks · Short✅ Verified
(a) The vertices of a rhombus ABCD are A(-3, -4), B(5, -3), C(1, 4) and D(-7, 3). Find the length of both the diagonals. Hence, find area of the rhombus ABCD.
OR
(b) The line segment joining the points A(-5, 1) and B(7, 6) is trisected at the points P and Q such that P is nearer to A. If P lies on the line x + y = k, then find the value of k.
✅ Answer & Solution
**(a)** $$AC=\sqrt{(1-(-3))^2+(4-(-4))^2}=\sqrt{16+64}=\sqrt{80}=4\sqrt5$$ $$BD=\sqrt{(-7-5)^2+(3-(-3))^2}=\sqrt{144+36}=\sqrt{180}=6\sqrt5$$ $$\text{Area of rhombus}=\frac12\times d_1\times d_2=\frac12\times4\sqrt5\times6\sqrt5=\frac12\times120=60\ sq.\ units$$
**(b)** P divides AB in ratio 1:2 (since P is nearer to A, trisection point closer to A):
$$P=\left(\frac{1(7)+2(-5)}{3},\frac{1(6)+2(1)}{3}\right)=\left(\frac{-3}{3},\frac{8}{3}\right)=(-1,\frac83)$$
Since P lies on $x+y=k$: $$k=-1+\frac83=\frac53$$
