Question

P (2, 3) and Q (6, 10) are two points on the Cartesian plane.

Answer 1

Answer:

length = square root of ((2-6)^2)+((3-10)^2) = square root (of) 16+49

square root of 65 =  8.062257748  = approx 8.1

midpoint = (x1+x2)/2  ,  (y1+y2)/2

= 2+6  /2  ,  3+10  /2

=8/2 , 13/2

= 4, 6.5 (midpoint)

equation= y=mx+c

m(gradient) = (y2-y1)/(x2-x1) or (y1-y2)/(x1-x2)

= so,  (3-10)/(2-6)  (y1-y2)/(x1-x2)- i normally use this one

= -7/-4 = 1.75

so y=1.75x+c

plug in the values from any one of them (2,3) or (6,10)

to get

for example:  3=1.75*2+c

3=3.5+c

-0.5=c

now we get the equation y=1.75x+(-0.5)

you always have to have a plus( +(-0.5)) since the equation is y=mx+c

Answer 2

Step-by-step explanation:

Given:

$(x_1, \: y_1 ) =(2, \:3)\:\:\&\:\: (x_2, \: y_2) =(6, \:10)$

$l(PQ) = \sqrt{ {(x_1 - x_2)}^{2} + {(y_1 - y_2)}^{2}} \\ \\ = \sqrt{ {(2 - 6)}^{2} + {(3 - 10)}^{2}} \\ \\ = \sqrt{ {( - 4)}^{2} + {( - 7)}^{2}} \\ \\ = \sqrt{16 + 49} \\ \\ = \sqrt{65} \\ \\ \therefore \purple{ \boxed{ l(PQ) = 8.06 \: units}}$

Let S be the mid-point of PQ.

$\therefore S = \{\frac{x_1+x_2}{2}, \:\:\frac{y_1+y_2}{2}\} \\\\= \{\frac{2 +6}{2}, \:\:\frac{3+10}{2}\} \\\\= \{\frac{8}{2}, \:\:\frac{13}{2}\} \\\\\therefore S= \{4, \:\:6.5\}$

Equation of line PQ is given as:

$\frac{y-y_1}{y_1 - y_2}=\frac{x-x_1}{x_1 - x_2} \\\\\therefore \frac{y-3}{3 - 10}=\frac{x-2}{2 - 6} \\\\\therefore \frac{y-3}{-7}=\frac{x-2}{-4} \\\\\therefore \frac{y-3}{7}=\frac{x-2}{4} \\\\\therefore 4(y-3)=7(x-2)\\\\\therefore 4y-12=7x-14\\\\\therefore 7x-14 - 4y + 12=0\\\\\red{\boxed {\therefore 7x-4y - 2 =0}}$