The area of a triangle is 5 sq. units. Two of its vertices are at $(2, 1)$ and $(3, -2)$. If the third vertex is $(\frac{7}{2} , y)$, find $y$.
Given:
The vertices of a triangle are $(2, 1), (3, -2)$ and $(\frac{7}{2}, y)$ and its area is 5 sq. units.
To do:
We have to find the value of $y$.
Solution:
Let $A(2, 1), B(3, -2)$ and $C(\frac{7}{2}, y)$ be the vertices of $\triangle ABC$.
We know that,
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle \( ABC=\frac{1}{2}[2(-2-y)+3(y-1)+\frac{7}{2}(1+2)] \)
\( \pm 5=\frac{1}{2}[-4-2y+3y-3+\frac{7}{2}(3)] \)
\( \pm 5(2)=(y-7+\frac{21}{2}) \)
\( \pm 10+7-\frac{21}{2}=y \)
\( y=\frac{2(17)-21}{2} \) or \( y=\frac{2(-3)-21}{2} \)
\( y=\frac{34-21}{2} \) or \( y=\frac{-6-21}{2} \)
\( y=\frac{13}{2} \) or \( y=\frac{-27}{2} \)
The value of $y$ is $\frac{13}{2}$ or $\frac{-27}{2}$.
Related Articles
- The area of a triangle is 5. Two of its vertices are $(2, 1)$ and $(3, -2)$. The third vertex lies on $y = x + 3$. Find the third vertex.
- Two vertices of a triangle are $(1, 2), (3, 5)$ and its centroid is at the origin. Find the coordinates of the third vertex.
- Find the third vertex of a triangle, if two of its vertices are at $(-3, 1)$ and $(0, -2)$ and the centroid is at the origin.
- Two vertices of an isosceles triangle are $(2, 0)$ and $(2, 5)$. Find the third vertex if the length of the equal sides is 3.
- $A (3, 2)$ and $B (-2, 1)$ are two vertices of a triangle ABC whose centroid $G$ has the coordinates $(\frac{5}{3}, −\frac{1}{3})$. Find the coordinates of the third vertex $C$ of the triangle.
- If $D (\frac{−1}{2}, \frac{5}{2}), E (7, 3)$ and $F (\frac{7}{2}, \frac{7}{2})$ are the mid-points of sides of $\triangle ABC$, find the area of $\triangle ABC$.
- If the vertices of a triangle are $(1, -3), (4, p)$ and $(-9, 7)$ and its area is 15 sq. units, find the value(s) of $p$.
- Which of the following is not true?A. $( y+\frac{1}{y})^2=( y-\frac{1}{y})^2+4$B. $( y-\frac{1}{y})^2+2=y^2+\frac{1}{y^2}$C. $( y+\frac{1}{y})^2-2=( y-\frac{1}{y})^2$D. $( y-\frac{1}{y})^2-( y+\frac{1}{y})^2=-(2)^2$
- \Find $(x +y) \div (x - y)$. if,(i) \( x=\frac{2}{3}, y=\frac{3}{2} \)(ii) \( x=\frac{2}{5}, y=\frac{1}{2} \)(iii) \( x=\frac{5}{4}, y=\frac{-1}{3} \)(iv) \( x=\frac{2}{7}, y=\frac{4}{3} \)(v) \( x=\frac{1}{4}, y=\frac{3}{2} \)
- (i) \( x^{2}-3 x+5-\frac{1}{2}\left(3 x^{2}-5 x+7\right) \)(ii) \( [5-3 x+2 y-(2 x-y)]-(3 x-7 y+9) \)(iii) \( \frac{11}{2} x^{2} y-\frac{9}{4} x y^{2}+\frac{1}{4} x y-\frac{1}{14} y^{2} x+\frac{1}{15} y x^{2}+ \) \( \frac{1}{2} x y \)(iv) \( \left(\frac{1}{3} y^{2}-\frac{4}{7} y+11\right)-\left(\frac{1}{7} y-3+2 y^{2}\right)- \) \( \left(\frac{2}{7} y-\frac{2}{3} y^{2}+2\right) \)(v) \( -\frac{1}{2} a^{2} b^{2} c+\frac{1}{3} a b^{2} c-\frac{1}{4} a b c^{2}-\frac{1}{5} c b^{2} a^{2}+ \) \( \frac{1}{6} c b^{2} a+\frac{1}{7} c^{2} a b+\frac{1}{8} c a^{2} b \).
- Find the value of y : $\frac{y^2 + 1}{y^2 - 1} = \frac{5}{4}$
- Add the following algebraic expressions(i) \( 3 a^{2} b,-4 a^{2} b, 9 a^{2} b \)(ii) \( \frac{2}{3} a, \frac{3}{5} a,-\frac{6}{5} a \)(iii) \( 4 x y^{2}-7 x^{2} y, 12 x^{2} y-6 x y^{2},-3 x^{2} y+5 x y^{2} \)(iv) \( \frac{3}{2} a-\frac{5}{4} b+\frac{2}{5} c, \frac{2}{3} a-\frac{7}{2} b+\frac{7}{2} c, \frac{5}{3} a+ \) \( \frac{5}{2} b-\frac{5}{4} c \)(v) \( \frac{11}{2} x y+\frac{12}{5} y+\frac{13}{7} x,-\frac{11}{2} y-\frac{12}{5} x-\frac{13}{7} x y \)(vi) \( \frac{7}{2} x^{3}-\frac{1}{2} x^{2}+\frac{5}{3}, \frac{3}{2} x^{3}+\frac{7}{4} x^{2}-x+\frac{1}{3} \) \( \frac{3}{2} x^{2}-\frac{5}{2} x-2 \)
- Find the area of the triangle whose vertices are:(i) $(2, 3), (-1, 0), (2, -4)$(ii) $(-5, -1), (3, -5), (5, 2)$
- An equilateral triangle has two vertices at the points $(3, 4)$, and $(-2, 3)$, find the coordinates of the third vertex.
- Subtract:(i) $-5xy$ from $12xy$(ii) $2a^2$ from $-7a^2$(iii) \( 2 a-b \) from \( 3 a-5 b \)(iv) \( 2 x^{3}-4 x^{2}+3 x+5 \) from \( 4 x^{3}+x^{2}+x+6 \)(v) \( \frac{2}{3} y^{3}-\frac{2}{7} y^{2}-5 \) from \( \frac{1}{3} y^{3}+\frac{5}{7} y^{2}+y-2 \)(vi) \( \frac{3}{2} x-\frac{5}{4} y-\frac{7}{2} z \) from \( \frac{2}{3} x+\frac{3}{2} y-\frac{4}{3} z \)(vii) \( x^{2} y-\frac{4}{5} x y^{2}+\frac{4}{3} x y \) from \( \frac{2}{3} x^{2} y+\frac{3}{2} x y^{2}- \) \( \frac{1}{3} x y \)(viii) \( \frac{a b}{7}-\frac{35}{3} b c+\frac{6}{5} a c \) from \( \frac{3}{5} b c-\frac{4}{5} a c \)
Kickstart Your Career
Get certified by completing the course
Get Started
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies