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Exercises · Q19

Q.Find the perpendicular distance of the point (2,3)(2, 3) from the line 4x+3y−10=04x + 3y - 10 = 0.

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Apply d=∣ax1+by1+c∣a2+b2d = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} with a=4a = 4, b=3b = 3, c=−10c = -10 and (x1,y1)=(2,3)(x_1, y_1) = (2, 3):

d=∣4(2)+3(3)−10∣42+32=∣8+9−10∣16+9=∣7∣25=75d = \dfrac{|4(2) + 3(3) - 10|}{\sqrt{4^2 + 3^2}} = \dfrac{|8 + 9 - 10|}{\sqrt{16 + 9}} = \dfrac{|7|}{\sqrt{25}} = \dfrac{7}{5} …

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