If the point of intersection of the lines x + 2y - 3 = 0 and x - 4y - 9 = 0 is 5 units away from the center of the circle having coordinates (x, 4), the value of x is -

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Solution

The correct option is E
5

The point of intersection of the lines x + 2y - 3 = 0 and x - 4y - 9 = 0 is (5,-1).

The point which is at a distance of 5 units from the point (5, -1) is (x, 4).

Distance formula = $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Solving we get x = 5.

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If the point of intersection of the lines x + 2y - 3 = 0 and x - 4y - 9 = 0 is 5 units away from the center of the circle having coordinates (x, 4), the value of x is -

The correct option is E 5 The point of intersection of the lines x + 2y - 3 = 0 and x - 4y - 9 = 0 is (5,-1). The point which is at a distance of 5 units from the point (5, -1) is (x, 4). Distance formula = √(x2−x1)2+(y2−y1)2 Solving we get x = 5.

The correct option is E
5

The point of intersection of the lines x + 2y - 3 = 0 and x - 4y - 9 = 0 is (5,-1).

The point which is at a distance of 5 units from the point (5, -1) is (x, 4).

Distance formula = $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Solving we get x = 5.