Find the distance of the point (3,−5)(3, -5)(3,−5) from the line 3x−4y−26=03x - 4y - 26 = 03x−4y−26=0.
Using formula: d=∣Ax1+By1+C∣A2+B2d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}d=A2+B2∣Ax1+By1+C∣ Substitute x1=3,y1=−5,A=3,B=−4,C=−26x_1=3, y_1=-5, A=3, B=-4, C=-26x1=3,y1=−5,A=3,B=−4,C=−26: d=∣3(3)−4(−5)−26∣32+(−4)2d = \frac{|3(3) - 4(-5) - 26|}{\sqrt{3^2 + (-4)^2}}d=32+(−4)2∣3(3)−4(−5)−26∣ d=∣9+20−26∣9+16d = \frac{|9 + 20 - 26|}{\sqrt{9 + 16}}d=9+16∣9+20−26∣ d=∣3∣25=35 unitsd = \frac{|3|}{\sqrt{25}} = \frac{3}{5} \text{ units}d=25∣3∣=53 units
Find the distance between the parallel lines 15x+8y−34=015x + 8y - 34 = 015x+8y−34=0 and 15x+8y+31=015x + 8y + 31 = 015x+8y+31=0.
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