Solution: According to the question, the points are (2, 3), (5, 7), and (-3, -1). The area of triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is as follows: = ½ [x1(y2 – y3) + x2(y3 -y1) +...

Solution: We have the equation x2 + xy – 3x + 2 = 0 The origin has been relocated from (0, 0) to  (p, q) As a result, any arbitrary point (x, y) is changed to (x + p, y + q). Therefore, the new...

(iii) xy – x – y + 1 = 0 (iv) xy – y2 – x + y = 0 Solution: (iii) xy – x – y + 1 = 0 We will replace the value of x by x + 1 and y by y + 1 Then, (x + 1) (y + 1) – (x + 1) – (y + 1) + 1 = 0 xy + x +...

(i) x2 + xy – 3x – y + 2 = 0 (ii) x2 – y2 – 2x + 2y = 0 Solution: (i) x2 + xy – 3x – y + 2 = 0 We will first substitute the value of x by x + 1 and y by y + 1. Then, te above-given ewuation becomes:...

Solution: The above-given equation \[\left( a-b \right)\left( {{x}^{2}}~+\text{ }{{y}^{2}} \right)-2abx\text{ }=\text{ }0\] can be rewritten into a new equation by replacing x by [X + ab / (a-b)]...

Solution: We have the equation: \[{{\left( x-a \right)}^{~2}}+{{\left( y-b \right)}^{~2}}~=\text{ }{{r}^{2}}\] The above-given equation (x – a)2 + (y – b)2 = r2 can be re-written into a new equation...