
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
162.9k+ views
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.
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.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NEET 2025 – Every New Update You Need to Know

Verb Forms Guide: V1, V2, V3, V4, V5 Explained

NEET Total Marks 2025

1 Billion in Rupees
