CBSE Class 10 Mathematics Linear Equations Assignment Set 14

LINEAR EQUATION

Q1. Draw the graphs of the following equations
a) \(x = 5\)
b) \(x = -3\)
c) \(y = 7\)
d) \(y = -5\)
Answer:
a) The equation \(x = 5\) represents a vertical line parallel to the y-axis at a distance of 5 units to the right of the y-axis. Every point on this line has an x-coordinate of 5.
b) The equation \(x = -3\) represents a vertical line parallel to the y-axis at a distance of 3 units to the left of the y-axis. Every point on this line has an x-coordinate of -3.
c) The equation \(y = 7\) represents a horizontal line parallel to the x-axis at a distance of 7 units above the x-axis. Every point on this line has a y-coordinate of 7.
d) The equation \(y = -5\) represents a horizontal line parallel to the x-axis at a distance of 5 units below the x-axis. Every point on this line has a y-coordinate of -5.
In simple words: Equations with only x form vertical lines, and equations with only y form horizontal lines.

Exam Tip: Remember that \(x = k\) is a vertical line passing through \((k, 0)\) and \(y = k\) is a horizontal line passing through \((0, k)\).

Q2. Draw the graphs of
a) \(y = x\)
b) \(y = -x\)
c) \(y = 3x\)
Answer:
a) For \(y = x\), some points on the line are \((0, 0)\), \((1, 1)\), and \((-1, -1)\). Plot these points and join them with a straight line passing through the origin at a 45-degree angle.
b) For \(y = -x\), some points on the line are \((0, 0)\), \((1, -1)\), and \((-1, 1)\). Plot these points and join them with a straight line passing through the origin sloping downwards.
c) For \(y = 3x\), some points on the line are \((0, 0)\), \((1, 3)\), and \((-1, -3)\). Plot these points and join them to get a steeper line passing through the origin.
In simple words: Linear equations of the form \(y = mx\) always pass through the origin \((0, 0)\).

Exam Tip: Always plot at least three points to ensure accuracy when drawing linear graphs.

Q3. Draw the graphs of
a) \(2x + 3y = 0\)
b) \(y = x - 7\)
Answer:
a) For \(2x + 3y = 0\), we can write \(y = -\frac{2}{3}x\).
When \(x = 0\), \(y = 0\), giving point \((0, 0)\).
When \(x = 3\), \(y = -2\), giving point \((3, -2)\).
When \(x = -3\), \(y = 2\), giving point \((-3, 2)\).
Plot \((0, 0)\), \((3, -2)\), and \((-3, 2)\) and draw a line through them.
b) For \(y = x - 7\):
When \(x = 0\), \(y = -7\), giving point \((0, -7)\).
When \(x = 7\), \(y = 0\), giving point \((7, 0)\).
When \(x = 4\), \(y = -3\), giving point \((4, -3)\).
Plot \((0, -7)\), \((7, 0)\), and \((4, -3)\) and draw a straight line through them.
In simple words: Find at least three coordinate points by substituting values of x, plot them, and draw a straight line passing through them.

Exam Tip: Choosing multiples of the denominator for x makes the calculation of y easier in fractional equations.

Q4. Draw the graphs of \(5x - 12y = 30\) and \(12x + 5y = 0\) and show that they are at right angles.
Answer:
For line 1: \(5x - 12y = 30\)
When \(x = 0\), \(-12y = 30 \implies y = -2.5\), point is \((0, -2.5)\).
When \(x = 6\), \(30 - 12y = 30 \implies y = 0\), point is \((6, 0)\).
Slope of line 1 (\(m_1\)) = \(\frac{5}{12}\).

For line 2: \(12x + 5y = 0\)
When \(x = 0\), \(y = 0\), point is \((0, 0)\).
When \(x = -5\), \(12(-5) + 5y = 0 \implies 5y = 60 \implies y = 12\), point is \((-5, 12)\).
Slope of line 2 (\(m_2\)) = \(-\frac{12}{5}\).

To show lines are at right angles, find the product of their slopes:
\(m_1 \times m_2 = \left(\frac{5}{12}\right) \times \left(-\frac{12}{5}\right) = -1\)
Since the product of the slopes is -1, the two lines are perpendicular to each other (at right angles).
In simple words: Two lines are perpendicular if the product of their slopes equals -1.

Exam Tip: Rearrange equations into slope-intercept form \(y = mx + c\) to easily identify the slopes \(m_1\) and \(m_2\).

Q5. Plot the graph of
a) \(4x + 3y = 12\)
b) \(6x - 5y = 30\)
and find the intercepts made by the line on the coordinate axes.
Answer:
a) For \(4x + 3y = 12\):
X-intercept: Put \(y = 0\), \(4x = 12 \implies x = 3\). Point is \((3, 0)\).
Y-intercept: Put \(x = 0\), \(3y = 12 \implies y = 4\). Point is \((0, 4)\).
Graph passes through \((3, 0)\) and \((0, 4)\).
X-intercept = 3 units, Y-intercept = 4 units.

b) For \(6x - 5y = 30\):
X-intercept: Put \(y = 0\), \(6x = 30 \implies x = 5\). Point is \((5, 0)\).
Y-intercept: Put \(x = 0\), \(-5y = 30 \implies y = -6\). Point is \((0, -6)\).
Graph passes through \((5, 0)\) and \((0, -6)\).
X-intercept = 5 units, Y-intercept = -6 units.
In simple words: The x-intercept is found by putting \(y = 0\), and the y-intercept is found by putting \(x = 0\).

Exam Tip: Finding intercepts is the fastest method to sketch a linear equation graph without needing many calculation points.

Q6. Plot the points \((12, 4)\) and \((-3, -5)\) and find if the straight line joining them passes through the point \((8, 2)\).
Answer:
First, find the equation of the line passing through \((x_1, y_1) = (12, 4)\) and \((x_2, y_2) = (-3, -5)\).
Slope \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-5 - 4}{-3 - 12} = \frac{-9}{-15} = \frac{3}{5}\).

Using point-slope form with point \((12, 4)\):
\(y - 4 = \frac{3}{5}(x - 12)\)
\(5(y - 4) = 3(x - 12)\)
\(5y - 20 = 3x - 36\)
\(3x - 5y - 16 = 0\)

Now, check if the point \((8, 2)\) satisfies this equation by substituting \(x = 8\) and \(y = 2\):
\(3(8) - 5(2) - 16 = 24 - 10 - 16 = -2 \neq 0\)
Since the point does not satisfy the equation, the straight line does not pass through \((8, 2)\).
In simple words: Substitute the coordinates of the test point into the equation of the line. If it balances to zero, it lies on the line.

Exam Tip: You can also check collinearity by verifying if the slope between the first point and the test point equals the slope of the line.

Q7. Plot \(2x - 3y = 12\); \(x = 1\) and \(y = 0\). Find the vertices of the triangle formed by these lines
Answer:
We have three lines forming a triangle:
Line 1: \(2x - 3y = 12\)
Line 2: \(x = 1\)
Line 3: \(y = 0\) (which is the x-axis)

Find the intersection points (vertices) by solving pairs of equations simultaneously:
1. Intersection of Line 2 (\(x = 1\)) and Line 3 (\(\mathbf{y = 0}\)):
Vertex 1 = \((1, 0)\)

2. Intersection of Line 1 (\(2x - 3y = 12\)) and Line 3 (\(y = 0\)):
Substitute \(y = 0\) into Line 1: \(2x - 3(0) = 12 \implies 2x = 12 \implies x = 6\).
Vertex 2 = \((6, 0)\)

3. Intersection of Line 1 (\(2x - 3y = 12\)) and Line 2 (\(x = 1\)):
Substitute \(x = 1\) into Line 1: \(2(1) - 3y = 12 \implies 2 - 3y = 12 \implies -3y = 10 \implies y = -\frac{10}{3}\).
Vertex 3 = \(\left(1, -\frac{10}{3}\right)\)

Vertices of the triangle are \((1, 0)\), \(\left(1, -\frac{10}{3}\right)\), and \((6, 0)\).
In simple words: The vertices of a triangle formed by intersecting lines are the points where each pair of lines intersects.

Exam Tip: Match your final vertices with the answer key provided: \((1, 0), (1, -\frac{10}{3}), (6, 0)\).

Q8. Draw graphs of \(x - y + 1 = 0\) and \(3x + 2y = 12\). Find the coordinates of the triangle formed by these lines and x-axis. Find its Area also?
Answer:
Line 1: \(x - y + 1 = 0 \implies y = x + 1\)
- When \(x = 0\), \(y = 1\) -> \((0, 1)\)
- When \(x = -1\), \(y = 0\) -> \((-1, 0)\)
- When \(x = 2\), \(y = 3\) -> \((2, 3)\)

Line 2: \(3x + 2y = 12\)
- When \(x = 0\), \(2y = 12 \implies y = 6\) -> \((0, 6)\)
- When \(x = 4\), \(3(4) + 2y = 12 \implies 2y = 0 \implies y = 0\) -> \((4, 0)\)

Line 3: x-axis (\(y = 0\))

Find vertices of the triangle:
1. Intersection of Line 1 and Line 2:
From Line 1, \(y = x + 1\). Substitute into Line 2:
\(3x + 2(x + 1) = 12\)
\(3x + 2x + 2 = 12\)
\(5x = 10 \implies x = 2\)
\(y = 2 + 1 = 3\)
First Vertex = \((2, 3)\)

2. Intersection of Line 1 and x-axis (\(y = 0\)):
\(x - 0 + 1 = 0 \implies x = -1\)
Second Vertex = \((-1, 0)\)

3. Intersection of Line 2 and x-axis (\(y = 0\)):
\(3x + 2(0) = 12 \implies 3x = 12 \implies x = 4\)
Third Vertex = \((4, 0)\)

Coordinates of the vertices of the triangle are \((2, 3)\), \((-1, 0)\), and \((4, 0)\).

To find the Area of the triangle:
Base along the x-axis = Distance from \(x = -1\) to \(x = 4\) = \(4 - (-1) = 5\) units.
Height = Y-coordinate of vertex \((2, 3)\) = 3 units.
Area = \(\frac{1}{2} \times \text{Base} \times \text{Height}\)
Area = \(\frac{1}{2} \times 5 \times 3 = \frac{15}{2} = 7.5\) square units.
In simple words: The base lies along the x-axis between the two x-intercepts, and the height is the y-coordinate of the intersection point of the two lines.

Exam Tip: Whenever a triangle is bounded by the x-axis, always take the length of the base as the difference between the two x-intercepts.