(C) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Given $A(-4, -3)$ and $B(6, a)

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(C) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Given $A(-4, -3)$ and $B(6, a)$ with distance $d = 10$.$10 = \sqrt{(6 - (-4))^2 + (a - (-3))^2}$$10 = \sqrt{(10)^2 + (a + 3)^2}$Squaring both sides:$100 = 100 + (a + 3)^2$$(a + 3)^2 = 0$$a + 3 = 0$$a = -3$

Class 10 Mathematics Coordinate Geometry Mix Examples - Coordinate Geometry

Medium

If the distance between $A(-4, -3)$ and $B(6, a)$ is $10$,then $a = \ldots \ldots \ldots . . .$

A
$4$
B
$3$
C
$-3$
D
$-4$

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Mix Examples - Coordinate Geometry

If the points are $(12, 10)$ and $(0, 8)$,then the midpoint of the line segment joining these two points is $\ldots \ldots \ldots \ldots$

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Prove that $A (1, 7)$,$B (2, 4)$,and $C (5, 5)$ are the vertices of an isosceles right triangle.

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If the distance between the points $(2, -2)$ and $(-1, x)$ is $5$,one of the values of $x$ is

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The distance of point $A(a \cos \theta, a \sin \theta)$ from the origin is $\ldots \ldots \ldots \ldots$ $(a \in R^{+})$

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$Y-$ axis divides the line segment joining $A(-3, -4)$ and $B(1, -2)$ from $A$ in the ratio............

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If the distance between $A(-4, -3)$ and $B(6, a)$ is $10$,then $a = \ldots \ldots \ldots . . .$

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If the points are $(12, 10)$ and $(0, 8)$,then the midpoint of the line segment joining these two points is $\ldots \ldots \ldots \ldots$

Prove that $A (1, 7)$,$B (2, 4)$,and $C (5, 5)$ are the vertices of an isosceles right triangle.

If the distance between the points $(2, -2)$ and $(-1, x)$ is $5$,one of the values of $x$ is

The distance of point $A(a \cos \theta, a \sin \theta)$ from the origin is $\ldots \ldots \ldots \ldots$ $(a \in R^{+})$

$Y-$ axis divides the line segment joining $A(-3, -4)$ and $B(1, -2)$ from $A$ in the ratio............

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