Question

If the point A is symmetric to the point B(4, -1) with respect to the bisector of first quadrant then AB^2​

Answer 1

Step-by-step explanation:

Given If the point A is symmetric to the point B(4, -1) with respect to the bisector of first quadrant then AB^2

• Bisector of first quadrant will be y = x
• If we plot a graph, and the line y = x then the slope m = 1
• Now the point is given as B(4, - 1)
• So if connect the line from B(4,-1) to  point A (a,b) and if the slope is m2 then m1 x m2 = -1
• Now there is a midpoint of line AB and the slope intersecting and let it be C, then it will be  
• c(4 + a/2 , b – ½)
• Also b – 1 / 2 = 4 + a / 2
•  Or b – 1 = 4 + a
•  Or a – b = - 5
• Now both the slopes are equal to – 1
• So (b + 1 / a – 4) x 1 = - 1
•    Or b + 1 = 4 – a
•     Or a + b = 3
•      Or a – b = - 5
•    So 2a = - 2
•        Or a = - 1
• So b = - 1 + 5
• Or b = 4
• Now we have the points A (- 1, 4)  and B (4, -1)
• Now we need to find the distance of AB
• So AB = √(-1 – 4)^2 + (4 + 1)^2
•         = √(-5)^2 + 5^2
•         = √50
•         = 5 √2  units
• Or AB^2 = (5√2)^2
•              = 50 units

Reference link will be

https://brainly.in/question/4238340