Find the value of a,if the distance between t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to the distance formula,the distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 -

Vedclass · Accepted Answer

$-3$

Vedclass · Answer

(B) According to the distance formula,the distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Given points are $A(-3, -14)$ and $B(a, -5)$ with distance $d = 9$.Substituting the values into the formula:$9 = \sqrt{(a - (-3))^2 + (-5 - (-14))^2}$$9 = \sqrt{(a + 3)^2 + (-5 + 14)^2}$$9 = \sqrt{(a + 3)^2 + (9)^2}$Squaring both sides:$81 = (a + 3)^2 + 81$Subtracting $81$ from both sides:$(a + 3)^2 = 0$Taking the square root:$a + 3 = 0$$a = -3$.Thus,the required value of $a$ is $-3$.

Find the value of $a$,if the distance between the points $A(-3, -14)$ and $B(a, -5)$ is $9$ units.

Similar Questions

If $P (9a-2, -b)$ divides the line segment joining $A (3a+1, -3)$ and $B (8a, 5)$ in the ratio $3:1$,find the values of $a$ and $b$.

State whether the following statement is true or false. Justify your answer.
Point $P(0, 2)$ is the point of intersection of the $y$-axis and the perpendicular bisector of the line segment joining the points $A(-1, 1)$ and $B(3, 3)$.

If $A(0, 0)$,$B(2, 0)$,$C(2, 2)$,and $D(0, 2)$,then $\square ABCD$ is a $\ldots \ldots \ldots$

The area of a triangle having vertices $A(3, 0)$,$B(0, 3)$,and $C(3, 3)$ is:

If $A(2, 9)$,$B(-2, 1)$,and $C(6, 3)$ are vertices of a triangle and the area of $\Delta ABC$ is $28$,find the length of the altitude drawn from $A$ to the side $\overline{BC}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module