Question

Find the value of y, if the distance travelled between the peints (2, y) and (-4, 3) is 10..

Answer 1

Answer: y=11.

Step-by-step explanation:

$(2;y_1)\ \ \ \ (-4;3)\ \ \ \ \ L=10\ \ \ \ \ y_1 > 0\ \ \ \ y_1=?\\The\ distance\ travelled\ between\ the\ peints:\\L=\sqrt{(x_2-x_1)^2+(y_2-y_1)}\\ 10=\sqrt{(-4-2)^2+(3-y_1)^2}\\ 10=\sqrt{(-6)^2+3^2-2*3*y_1+y_1^2}\\ 10=\sqrt{36+9-6y_1+y_1^2} \\10=\sqrt{45-6y_1+y_1^2}\\10*10= \sqrt{45-6y_1+y_1^2}*\sqrt{45-6y_1+y_1^2}\\100=45-6y_1+y^2_1$

$y^2-6y_1+45=100\\y_1^2-6y_1-55=0\\y^2_1-6y_1-5y+5y-55=0\\y^2_1-11y_1+5y_1-55=0\\y_1*(y_1-11)+5*(y_1-11)=0\\(y_1-11)*(y_1+5)=0\\y_1-11=0\\y_1=11.\\y_1+5=0\\y_1=-5\notin\ (y_1 > 0).$