Question

Two lines are drawn through \[\left( {3,4} \right)\], each of which makes an angle of \[45^\circ \] with the line \[x - y = 2\], then what is the area of the triangle formed by these lines?
A. \[9\]
B. \[\dfrac{9}{2}\]
C. \[2\]
D. \[\dfrac{2}{9}\]

Answer 1

Hint: Find the slope of the given line and identify the triangle is right-angled. Find the equations of the other two sides. One of the vertices is given. Find the other two vertices by solving the equations and hence find the area of the triangle.

Formula used
Slope-intercept form of an equation of a line is \[y = mx + c\], where \[m\] is slope and \[c\] is \[y\]-intercept.
If \[\theta \] be the angle made by a line with \[ + ve\] \[x\]-axis, then \[m = \tan \theta \]
The Sum of three angles of a triangle is \[180^\circ \]
Area of a right-angled triangle is \[A = \dfrac{1}{2} \times b \times h\] square units.

Complete step by step solution
The given line is \[x - y = 2\]
\[ \Rightarrow y = x - 2 - - - - - \left( i \right)\]
Comparing it with the line \[y = mx + c\], we get
Slope of the given line is \[m = 1\]
If \[\theta \] be the angle made by the line with \[ + ve\] \[x\]-axis, then \[\tan \theta = 1\]\[ \Rightarrow \theta = 45^\circ \]
So, two angles of the triangle are \[45^\circ \] and the third angle is \[180^\circ - \left( {45^\circ + 45^\circ } \right) = 180^\circ - 90^\circ = 90^\circ \]
Hence, it is a right-angled triangle.
It is clear that one of these lines is parallel to the \[x\]-axis and the other is parallel to the \[y\]-axis.
Since these lines passes through the point \[\left( {3,4} \right)\].
So, equation of one line is \[x = 3 - - - - - \left( {ii} \right)\]
and equation of the other line is \[y = 4 - - - - - \left( {iii} \right)\]
Let us find the other two vertices of the triangle by solving the equations of the line.
Putting \[x = 3\] from equation \[\left( {ii} \right)\] in equation \[\left( i \right)\], we get
\[y = 3 - 2 = 1\]
So, the vertex which is the point of intersection of the lines \[\left( i \right)\] and \[\left( {ii} \right)\] is \[\left( {3,1} \right)\]
Putting \[y = 4\] from equation \[\left( {iii} \right)\] in equation \[\left( i \right)\], we get
\[x = 2 + 4 = 6\]
So, the vertex which is the point of intersection of the lines \[\left( i \right)\] and \[\left( {iii} \right)\] is \[\left( {6,4} \right)\]
\[\therefore \]The vertices of the triangle are \[A\left( {3,4} \right)\], \[B\left( {3,1} \right)\] and \[C\left( {6,4} \right)\]
Length of the height of the triangle \[ABC\] is \[b = AB = 4 - 1 = 3\] and length of the base is \[h = BC = 6 - 3 = 3\]
\[\therefore \]Area of the triangle \[ABC\] is \[\dfrac{1}{2} \times 3 \times 3 = \dfrac{9}{2}\] square units.

Hence option B is correct.

Note: The points of intersection of the lines are the vertices of the triangle, which are obtained by solving the equations of the lines. The formula used here is applicable only for a right-angled triangle and for such a triangle whose base and height are given or can be found out.