How to answer this questions: coordinates of the midpoints and finding the equations of lines AC?

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1 Answer
Jan 30, 2018

"see explanation"see explanation

Explanation:

(a)(a)

"given "A(x_1,y_1)" and "B(x_2,y_2)given A(x1,y1) and B(x2,y2)

"then the coordinates of the midpoint (M) of AB"then the coordinates of the midpoint (M) of AB

•color(white)(x)M=[1/2(x_1+x_2),1/2(y_1+y_2)]xM=[12(x1+x2),12(y1+y2)]

rArrM=[1/2(9+5),1/2(10+0)]=(7,5)M=[12(9+5),12(10+0)]=(7,5)

rArrN=[1/2(9+13),1/2(10+0)]=(11,5)N=[12(9+13),12(10+0)]=(11,5)

(b)(b)

"equations of lines AC and BC"equations of lines AC and BC

"the equation of a line in "color(blue)"slope-intercept form"the equation of a line in slope-intercept form is.

•color(white)(x)y=mx+bxy=mx+b

"where m is the slope and b the y-intercept"where m is the slope and b the y-intercept

"calculate the slope m using the "color(blue)"gradient formula"calculate the slope m using the gradient formula

•color(white)(x)m=(y_2-y_1)/(x_2-x_1)xm=y2y1x2x1

rArrm_(AC)=(10-0)/(9-5)=10/4=5/2mAC=10095=104=52

rArry=5/2x+blarrcolor(blue)"partial equation"y=52x+bpartial equation

"to find b substitute the coordinates of either A or C"to find b substitute the coordinates of either A or C
"into the partial equation"into the partial equation

"using "A(5,0)" then"using A(5,0) then

0=25/2+brArrb=-25/20=252+bb=252

rArry=5/2x-25/2larrcolor(blue)"equation of AC"y=52x252equation of AC

"Similarly for the equation of BC"Similarly for the equation of BC

m_(BC)=(10-0)/(9-13)=10/(-4)=-5/2mBC=100913=104=52

rArry=-5/2x+by=52x+b

"using "B(13,0)" then"using B(13,0) then

0=-65/2+brArrb=65/20=652+bb=652

rArry=-5/2x+65/2larrcolor(blue)"equation of BC"y=52x+652equation of BC