How to answer this questions: coordinates of the midpoints and finding the equations of lines AC?
1 Answer
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+b∙xy=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=y2−y1x2−x1
rArrm_(AC)=(10-0)/(9-5)=10/4=5/2⇒mAC=10−09−5=104=52
rArry=5/2x+blarrcolor(blue)"partial equation"⇒y=52x+b←partial 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+b⇒b=−252
rArry=5/2x-25/2larrcolor(blue)"equation of AC"⇒y=52x−252←equation of AC
"Similarly for the equation of BC"Similarly for the equation of BC
m_(BC)=(10-0)/(9-13)=10/(-4)=-5/2mBC=10−09−13=10−4=−52
rArry=-5/2x+b⇒y=−52x+b
"using "B(13,0)" then"using B(13,0) then
0=-65/2+brArrb=65/20=−652+b⇒b=652
rArry=-5/2x+65/2larrcolor(blue)"equation of BC"⇒y=−52x+652←equation of BC