Question
Find distance $CD$ where $C(– 3a, a), D(a, – 2a)$
✓
Answer
Let $C\left(x_1, y_1\right)$ and $D\left(x_2, y_2\right)$ be the given points
$\therefore x_1=-3 a, y_1=a, x_2=a, y_2=-2 a$
By distance formula,
$ d(C, D)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$=\sqrt{[a-(-3 a)]^2+(-2 a-a)^2}$
$=\sqrt{(a+3 a)^2+(-2 a-a)^2}$
$=\sqrt{(4 a)^2+(-3 a)^2}$
$=\sqrt{16 a^2+9 a^2}$
$=\sqrt{25 a^2}$
$\therefore d(C, D)=5 a \text { units } $
$\therefore x_1=-3 a, y_1=a, x_2=a, y_2=-2 a$
By distance formula,
$ d(C, D)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$=\sqrt{[a-(-3 a)]^2+(-2 a-a)^2}$
$=\sqrt{(a+3 a)^2+(-2 a-a)^2}$
$=\sqrt{(4 a)^2+(-3 a)^2}$
$=\sqrt{16 a^2+9 a^2}$
$=\sqrt{25 a^2}$
$\therefore d(C, D)=5 a \text { units } $
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Show that the points A(1, 2), B(1, 6), C(1 + 2 √3, 4) are vertices of an equilateral triangle.
In the given figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.
Find the $37^{th}$ term of the AP $6,7\frac{3}{4},9\frac{1}{2},11\frac{1}{4},\ ...$
→In figure 3.76, seg PQ is a diameter of a circle with centre O. R is any point on the circle. seg RS⊥seg PQ.
Prove that, SR is the geometric mean of PS and SQ. [That is, SR² = PS ×SQ]
Prove that, SR is the geometric mean of PS and SQ. [That is, SR² = PS ×SQ]
Write an A.P. whose first term is a and common difference is d in each of the following.
$a = – 1.25, d = 3$
→$a = – 1.25, d = 3$
Evaluate the following:
$\frac{5\cos^260^\circ+4\sec^230^\circ-\tan^245^\circ}{\sin^230^\circ+\cos^230^\circ}$
→$\frac{5\cos^260^\circ+4\sec^230^\circ-\tan^245^\circ}{\sin^230^\circ+\cos^230^\circ}$
If $\theta=30^\circ,$ verify that.
$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$
→$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$
Prove the following trigonometric identities.
$\text{sin}^2\text{A}+\frac{1}{1+\tan^2\text{A}}=1$
→$\text{sin}^2\text{A}+\frac{1}{1+\tan^2\text{A}}=1$
If $\sin\theta=\frac{1}{3},$ then find is the value of $2\cot^2\theta+2$.
→Solve the following quadratic equation by factorization.
2m (m-24) = 50
2m (m-24) = 50