ABC is on isosceles triangle of
25/6 square units. If the
coordinates of the base are
B(1,3) ond C(-2,7), then the<

2 thoughts on “ABC is on isosceles triangle of<br />25/6 square units. If the<br />coordinates of the base are<br />B(1,3) ond C(-2,7), then the<”

1. Answer:

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2. Step-by-step explanation:

   Find the length of the base:

   Given that, the coordinates of the base are [tex]B (1, 3)[/tex] and [tex]C (- 2, 7)[/tex].

   Thus the length of [tex]BC[/tex]

   [tex]=\sqrt{(1+2)^{2}+(3-7)^{2}}[/tex] units

   [tex]=\sqrt{3^{2}+4^{2}}[/tex] units

   [tex]=\sqrt{9+16}[/tex] units

   [tex]=\sqrt{25}[/tex] units

   [tex]5[/tex] units

   Find the length of the height:

   Since the area of the given isosceles triangle [tex]ABC[/tex] is [tex]\frac{25}{6}[/tex] square units, the length of the height [tex]AD[/tex] (drawn) be

   [tex]\quad 2\times\frac{area\:of\:\Delta ABC}{length\:of\:the\:base}[/tex]

   [tex]=2\times\frac{\frac{25}{6}}{5}[/tex] units

   [tex]=2\times\frac{25}{6\times 5}[/tex] units

   [tex]=\frac{50}{30}[/tex] units

   [tex]=\frac{5}{3}[/tex] units

   Find the equation of the base:

   The equation of the base [tex]BC[/tex] is

   [tex]\quad \frac{y-7}{7-3}=\frac{x+2}{-2-1}[/tex]

   [tex]\Rightarrow \frac{y-7}{4}=\frac{x+2}{-3}[/tex]

   [tex]\Rightarrow 4x+8=-3y+21[/tex]

   [tex]\Rightarrow 4x+3y=13[/tex] ___(1)

   Find the mid-point of the base:

   The coordinates of the mid-point [tex]D[/tex] of the base [tex]BC[/tex] are

   [tex]\quad (\frac{1-2}{2},\frac{3+7}{2})[/tex] i.e., [tex](-\frac{1}{2},5)[/tex]

   Find the equation of the height:

   Since [tex]AD[/tex] is perpendicular to [tex]BC[/tex], let the equation of [tex]AD[/tex] be

   [tex]\quad 3x-4y=k[/tex] where [tex]k[/tex] is constant

   Since [tex]AD[/tex] pαsses through the point [tex]D(-\frac{1}{2},5)[/tex],

   [tex]\quad 3(-\frac{1}{2})-4(5)=k[/tex]

   [tex]\Rightarrow k=-\frac{43}{2}[/tex]

   Thus the equation of [tex]AD[/tex] is

   [tex]\quad 3x-4y=-\frac{43}{2}[/tex]

   [tex]\Rightarrow 6x-8y=-43[/tex] ___ (2)

   Consider the vertex [tex]A[/tex]:

   Let the coordinates of [tex]A[/tex] be [tex](m,n)[/tex].

   Satisfy equation no. (2) with [tex]A[/tex]:

   Since [tex]AD[/tex] is satisfied by [tex]A(m,n)[/tex],

   [tex]\quad 6m-8n=-43[/tex] ___ (3)

   Use the distance of [tex]A[/tex] from the base:

   Here the distance of [tex]A(m,n)[/tex] from the base [tex]BC:4x+3y=13[/tex] is [tex]\frac{5}{3}[/tex] units. Then

   [tex]\quad \frac{4m+3n-13}{\sqrt{4^{2}+3^{2}}}=\frac{5}{3}[/tex]

   [tex]\Rightarrow \frac{4m+3n-13}{5}=\frac{5}{3}[/tex]

   [tex]\Rightarrow 12m+9n-39=25[/tex]

   [tex]\Rightarrow 12m+9n=64[/tex] ___ (4)

   Find the value of [tex]m[/tex] and [tex]n[/tex]:

   We have two equations to solve,

   [tex]\quad 6m-8n=-43\quad\quad\times 2[/tex]

   [tex]\quad 12m+9n=64\quad\quad\times 1[/tex]

   [tex]\Rightarrow[/tex]

   [tex]\quad 12m-16n=-86[/tex]

   [tex]\quad 12m+9n=64[/tex]

   On subtraction, we get

   [tex]\quad 25n=150\Rightarrow n=6[/tex]

   Thus [tex]m=\frac{5}{6}[/tex]

   Answer:

   [tex]\therefore[/tex] the coordinates of [tex]A[/tex] are [tex](\frac{5}{6},6)[/tex].

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