Answer:
To find the equation of the line perpendicular to the line � = − 5 7 � + 3 y=− 7 5 x+3 and passing through the point ( 5 , − 2 ) (5,−2), you can follow these steps: First, determine the slope ( � m) of the original line � = − 5 7 � + 3 y=− 7 5 x+3. In this case, � = − 5 7 m=− 7 5 . The slope of a line perpendicular to the original line will be the negative reciprocal of the original slope. So, the slope ( � ′ m ′ ) of the perpendicular line is 7 5 5 7 . Now that you have the slope ( � ′ m ′ ) of the perpendicular line, you can use the point-slope form of a linear equation to find the equation of the perpendicular line. The point-slope form is: � − � 1 = � ′ ( � − � 1 ) y−y 1 =m ′ (x−x 1 ) where ( � 1 , � 1 ) (x 1 ,y 1 ) is the point through which the line passes. In this case, ( � 1 , � 1 ) = ( 5 , − 2 ) (x 1 ,y 1 )=(5,−2). Plugging in these values: � − ( − 2 ) = 7 5 ( � − 5 ) y−(−2)= 5 7 (x−5) Simplify: � + 2 = 7 5 ( � − 5 ) y+2= 5 7 (x−5) Now, let's isolate � y on the left side: � = 7 5 ( � − 5 ) − 2 y= 5 7 (x−5)−2 Now, distribute the 7 5 5 7 on the right side: � = 7 5 � − 7 − 2 y= 5 7 x−7−2 Combine the constants on the right side: � = 7 5 � − 9 y= 5 7 x−9 So, the equation of the line perpendicular to � = − 5 7 � + 3 y=− 7 5 x+3 and passing through the point ( 5 , − 2 ) (5,−2) is: � = 7 / 5 � − 9 y= x 7/5 −9 This is the equation in the � = � � + � y=mx+b form.