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Last updated at Sept. 6, 2021 by Teachoo
Example 18 Find the distance of the point (3, β5) from the line 3x β 4y β26 = 0. We know that distance (d) of a point (x1, y1) from a line Ax + By + C = 0 is d = |π΄π₯_1 + γπ΅π¦γ_2 + πΆ|/β(π΄^2 + π΅^2 ) Now, our equation is 3x β 4y β 26 = 0 The above equation is of the form Ax + By + C = 0 where A = 3, B = β4 , C = β26 & we have to find the distance of the point (3, β 5) from the line So, x1 = 3 , y1 = β5 Now finding distance d = |π΄π₯_1 + γπ΅π¦γ_2 + πΆ|/β(π΄^2 + π΅^2 ) Putting values = |3(3) + (β4)( β5) β 26|/β(32 + (β4)2) = |9 + 20 β 26|/β(9 + 16) = |29 β 26|/β25 = |3|/β(5 Γ 5) = |3|/5 = 3/5 β΄ Required distance = d = π/π units