Q1. Write the equations for the x-and y-axes.

Solution

The equation of the x-axis represents all points whose y-coordinate is zero because every point on the x-axis lies horizontally along \(y = 0\).

\[ \begin{aligned} y = 0 \end{aligned} \]

The equation of the y-axis represents all points whose x-coordinate is zero because every point on the y-axis lies vertically along \(x = 0\).

\[ \begin{aligned} x = 0 \end{aligned} \]

Hence, the equations of the x-axis and y-axis are \(y = 0\) and \(x = 0\) respectively

Q2. Passing through the point (– 4, 3) with slope \(\frac{1}{2}\)

Solution

The line passes through the point \((-4, 3)\) and has slope \(m = \frac{1}{2}\). Using the point–slope form of the equation of a line,

\[ \begin{aligned} y - y_1 &= m(x - x_1) \end{aligned} \]

Substituting \(x_1 = -4\), \(y_1 = 3\), and \(m = \frac{1}{2}\),

\[ \begin{aligned} y - 3 &= \frac{1}{2}\left(x - (-4)\right) \\ y - 3 &= \frac{1}{2}(x + 4) \end{aligned} \]

Multiplying both sides by \(2\) to remove the fraction,

\[ \begin{aligned} 2y - 6 &= x + 4 \\ 2y - x - 10 &= 0 \end{aligned} \]

Hence, the required equation of the line is \(2y - x - 10 = 0\)

Q3. Passing through (0, 0) with slope m.

Solution

The line passes through the origin \((0, 0)\) and has slope \(m\). Using the point–slope form of the equation of a line,

\[ \begin{aligned} y - y_0 &= m(x - x_0) \end{aligned} \]

Substituting \(x_0 = 0\) and \(y_0 = 0\),

\[ \begin{aligned} y - 0 &= m(x - 0) \\ y &= mx \end{aligned} \]

Hence, the equation of the required line is \(y = mx\)

Q4. Passing through \((2, 2\sqrt{3})\) and inclined with the x-axis at an angle of \(75^\circ\).

Solution

The line passes through the point \((2, 2\sqrt{3})\) and is inclined at an angle of \(75^\circ\) with the positive \(x\)-axis. The slope of the line is given by \(m = \tan \theta\).

\[ \begin{aligned} m &= \tan 75^\circ \\ &= \tan(45^\circ + 30^\circ) \\ &= \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} \\ &= \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} \\ &= \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \\ &= \frac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \\ &= \frac{3 + 2\sqrt{3} + 1}{2} \\ &= 2 + \sqrt{3} \end{aligned} \]

Using the point–slope form of the equation of a line,

\[ \begin{aligned} y - y_0 &= m(x - x_0) \\ y - 2\sqrt{3} &= (2 + \sqrt{3})(x - 2) \end{aligned} \]

Expanding and simplifying,

\[ \begin{aligned} y - 2\sqrt{3} &= (2 + \sqrt{3})x - 2(2 + \sqrt{3}) \\ y - 2\sqrt{3} &= (2 + \sqrt{3})x - 4 - 2\sqrt{3} \\ y &= (2 + \sqrt{3})x - 4 \end{aligned} \]

Hence, the required equation of the line is \(y = (2 + \sqrt{3})x - 4\)

Q5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.

Solution

The line intersects the x-axis at a distance of 3 units to the left of the origin, so the point of intersection on the x-axis is \((-3, 0)\). The slope of the line is given as \(m = -2\).

Using the point–slope form of the equation of a line,

\[ \begin{aligned} y - y_0 &= m(x - x_0) \\ y - 0 &= -2\left(x - (-3)\right) \\ y &= -2(x + 3) \\ y &= -2x - 6 \\ y + 2x + 6 &= 0 \end{aligned} \]

Hence, the equation of the required line is \(y + 2x + 6 = 0\)

Q6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of \(30^\circ\) with positive direction of the x-axis.

Solution

The line intersects the y-axis at a distance of 2 units above the origin, so the point of intersection on the y-axis is \((0, 2)\). The line makes an angle of \(30^\circ\) with the positive direction of the x-axis.

The slope of the line is given by \(m = \tan \theta\)

\[ \begin{aligned} m &= \tan 30^\circ \\ &= \frac{1}{\sqrt{3}} \end{aligned} \]

Using the point–slope form of the equation of a line,

\[ \begin{aligned} y - y_0 &= m(x - x_0) \\ y - 2 &= \frac{1}{\sqrt{3}}(x - 0) \\ y - 2 &= \frac{x}{\sqrt{3}} \\ \sqrt{3}y - 2\sqrt{3} &= x \\ \sqrt{3}y - x - 2\sqrt{3} &= 0 \end{aligned} \]

Hence, the equation of the required line is \(\sqrt{3}y - x - 2\sqrt{3} = 0\)

Q7. Passing through the points (–1, 1) and (2, – 4).

Solution

The line passes through the points \((-1, 1)\) and \((2, -4)\). First, we find the slope of the line using the slope formula.

\[ \begin{aligned} m &= \frac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{-4 - 1}{2 - (-1)} \\ &= \frac{-5}{3} \end{aligned} \]

Using the point–slope form of the equation of a line with point \((-1, 1)\),

\[ \begin{aligned} y - y_0 &= m(x - x_0) \\ y - 1 &= \frac{-5}{3}\left(x - (-1)\right) \\ y - 1 &= \frac{-5}{3}(x + 1) \end{aligned} \]

Multiplying both sides by \(3\) to remove the denominator and simplifying,

\[ \begin{aligned} 3y - 3 &= -5x - 5 \\ 3y + 5x - 3 + 5 &= 0 \\ 3y + 5x + 2 &= 0 \end{aligned} \]

Hence, the equation of the required line is \(3y + 5x + 2 = 0\)

Q8. The vertices of \(\mathrm{\Delta PQR}\) are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R.

Solution

The vertices of the triangle \(\Delta PQR\) are \(P(2, 1)\), \(Q(-2, 3)\), and \(R(4, 5)\). The median from vertex \(R\) passes through \(R\) and the midpoint of \(PQ\).

The midpoint of \(PQ\) is found using the section formula. Let the coordinates of the midpoint of \(PQ\) be \((x, y)\)

\[ \begin{aligned} x &= \frac{2 + (-2)}{2} = 0 \\ y &= \frac{1 + 3}{2} = 2 \end{aligned} \]

Therefore, the median passes through the points \((0, 2)\) and \((4, 5)\). Now we find the slope of the median

\[ \begin{aligned} m &= \frac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{5 - 2}{4 - 0} \\ &= \frac{3}{4} \end{aligned} \]

Using the point–slope form of the equation of a line through \((0, 2)\),

\[ \begin{aligned} y - y_0 &= m(x - x_0) \\ y - 2 &= \frac{3}{4}(x - 0) \\ 4y - 8 &= 3x \\ 4y - 3x - 8 &= 0 \end{aligned} \]

Hence, the equation of the median through vertex \(R\) is \(4y - 3x - 8 = 0\)

Q9. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

Solution

The required line passes through the point \((-3, 5)\) and is perpendicular to the line passing through the points \((2, 5)\) and \((-3, 6)\). Let the slope of the given line be \(m_2\) and the slope of the required line be \(m_1\)

\[ \begin{aligned} m_{2} &= \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ &= \frac{6 - 5}{-3 - 2} \\ &= \frac{1}{-5} \\ &= -\frac{1}{5} \end{aligned} \]

Since the two lines are perpendicular, the product of their slopes is \(-1\)

\[ \begin{aligned} m_{1} \cdot m_{2} &= -1 \\ m_{1} &= -\frac{1}{m_{2}} \\ m_{1} &= -\frac{1}{-\frac{1}{5}} \\ m_{1} &= 5 \end{aligned} \]

Now using the point–slope form of the equation of a line through \((-3, 5)\)

\[ \begin{aligned} y - y_{0} &= m(x - x_{0}) \\ y - 5 &= 5\left(x - (-3)\right) \\ y - 5 &= 5(x + 3) \\ y - 5 &= 5x + 15 \\ y - 5x - 20 &= 0 \end{aligned} \]

Hence, the equation of the required line is \(y - 5x - 20 = 0\)

Q10. A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.

Solution

The line segment is formed by the points \((1, 0)\) and \((2, 3)\). Let the point of intersection of the perpendicular line dividing the segment in the ratio \(1 : n\) be \((x, y)\)

By section formula

\[ \begin{aligned} x &= \frac{1 \cdot n + 2 \cdot 1}{n + 1} = \frac{n + 2}{n + 1} \\ y &= \frac{0 \cdot n + 3 \cdot 1}{n + 1} = \frac{3}{n + 1} \end{aligned} \]

Slope of the given line segment is

\[ \begin{aligned} M_{2} &= \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ &= \frac{3 - 0}{2 - 1} \\ &= 3 \end{aligned} \]

Since the required line is perpendicular to this line

\[ \begin{aligned} M_{1} \cdot M_{2} &= -1 \\ M_{1} &= -\frac{1}{3} \end{aligned} \]

Equation of the perpendicular line passing through \(\left(\frac{n+2}{n+1}, \frac{3}{n+1}\right)\)

\[ \begin{aligned} y - y_{0} &= m(x - x_{0}) \\ y - \frac{3}{n+1} &= -\frac{1}{3}\left(x - \frac{n+2}{n+1}\right) \\ 3(n+1)y - 3 &= -(n+1)x + (n+2) \\ 3(n+1)y + (n+1)x &= n + 11 \end{aligned} \]

Hence, the equation of the required line is \(3(n+1)y + (n+1)x = n + 11\)

Q11. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

Solution

Improve my solution (but don’t change structure), correct if anything wrong, for the given question in a 100% human-written style, return an improved solution in HTML + Bootstrap + MathJax in code block do not include header and body but only the solution part, avoid lists, instead use the

element and avoid element. Write code in a neat and clean format Question: Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3). Solution: line cuts equal intercepts on the coordinate axes and passes through (2,3) Let intercept be 'a' Equation of line $$\begin{aligned}\dfrac{x}{a}+\dfrac{y}{b}=1\\ a=b\\ x+y=a\end{aligned}$$ this line must satisfy (2,3) $$\begin{aligned}x+y=a\\ \Rightarrow a=2+3=5\end{aligned}$$ Hence Equation of line becomes $$\begin{aligned}\dfrac{x}{5}+\dfrac{y}{5}=1\\ x+y=5\end{aligned}$$ Avoid giving [web:] references. Do not put (.) period within aligned environment also do not repeat question and Solution heading. Include necessary reasons and explanation in the solution.

Q12. Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

Solution

The sum of the intercepts on the coordinate axes is \(9\). Let the x-intercept be \(a\), then the y-intercept will be \(9 - a\). Since the line passes through the point \((2, 2)\), we use the intercept form of the equation of a line.

\[ \frac{x}{a} + \frac{y}{9 - a} = 1 \]

Substituting the point \((2, 2)\) into the equation,

\[ \begin{aligned} \frac{2}{a} + \frac{2}{9 - a} &= 1 \\ 2(9 - a) + 2a &= a(9 - a) \\ 18 - 2a + 2a &= 9a - a^2 \\ 18 &= 9a - a^2 \\ a^2 - 9a + 18 &= 0 \\ a^2 - 6a - 3a + 18 &= 0 \\ a(a - 6) - 3(a - 6) &= 0 \\ (a - 6)(a - 3) &= 0 \\ a &= 6 \\ a &= 3 \end{aligned} \]

When \(a = 6\), the equation of the line is

\[ \begin{aligned} \frac{x}{6} + \frac{y}{3} &= 1 \\ x + 2y &= 6 \end{aligned} \]

When \(a = 3\), the equation of the line is

\[ \begin{aligned} \frac{x}{3} + \frac{y}{6} &= 1 \\ 2x + y &= 6 \end{aligned} \]

Hence, the required equations of the line are \(x + 2y = 6\) and \(2x + y = 6\)

Q13. Find equation of the line through the point (0, 2) making an angle \(\frac{2\pi}{3}\) with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

Solution

The line passes through the point \((0, 2)\) and makes an angle \(\frac{2\pi}{3}\) with the positive x-axis. The slope of the line is given by \(m = \tan \theta\)

\[ \begin{aligned} m &= \tan \frac{2\pi}{3} \\ &= \tan(180^\circ - 60^\circ) \\ &= -\tan 60^\circ \\ &= -\sqrt{3} \end{aligned} \]

Using the point–slope form of the equation of a line

\[ \begin{aligned} y - y_{0} &= m(x - x_{0}) \\ y - 2 &= -\sqrt{3}(x - 0) \\ y - 2 &= -\sqrt{3}x \\ y + \sqrt{3}x - 2 &= 0 \end{aligned} \]

The line parallel to this one has the same slope and crosses the y-axis at a distance of 2 units below the origin, so the point of intersection is \((0, -2)\)

Equation of the parallel line

\[ \begin{aligned} y - y_{0} &= m(x - x_{0}) \\ y - (-2) &= -\sqrt{3}(x - 0) \\ y + 2 &= -\sqrt{3}x \\ y + \sqrt{3}x + 2 &= 0 \end{aligned} \]

Hence, the required equations are \(y + \sqrt{3}x - 2 = 0\) and \(y + \sqrt{3}x + 2 = 0\)

Q14. The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.

Solution

The point of intersection of the perpendicular drawn from the origin to the given line is \((-2, 9)\). The slope of the perpendicular line joining the origin \((0, 0)\) and \((-2, 9)\) is

\[ \begin{aligned} m &= \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ &= \frac{9 - 0}{-2 - 0} \\ &= -\frac{9}{2} \end{aligned} \]

Let the slope of the required line be \(M\). Since the given line is perpendicular to this segment,

\[ \begin{aligned} M \cdot m &= -1 \\ M \cdot \left(-\frac{9}{2}\right) &= -1 \\ M &= \frac{2}{9} \end{aligned} \]

Equation of the line with slope \(\frac{2}{9}\) and passing through \((-2, 9)\)

\[ \begin{aligned} y - y_{0} &= m(x - x_{0}) \\ y - 9 &= \frac{2}{9}\left(x - (-2)\right) \\ y - 9 &= \frac{2}{9}(x + 2) \\ 9y - 81 &= 2x + 4 \\ 9y - 2x - 85 &= 0 \end{aligned} \]

Hence, the equation of the required line is \(9y - 2x - 85 = 0\)

Q15. The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L= 125.134 when C = 110, express L in terms of C.

Solution

Since the length \(L\) of the copper rod varies linearly with temperature \(C\), we assume a linear relation of the form

\[ \begin{aligned} L = mC + b \end{aligned} \]

From the given data, the two points on the line are \((20, 124.942)\) and \((110, 125.134)\). The slope \(m\) is

\[ \begin{aligned} m &= \frac{125.134 - 124.942}{110 - 20} \\ &= \frac{0.192}{90} \\ &= 0.002133\overline{3} \end{aligned} \]

Substituting \(m\) and the point \((20, 124.942)\) into \(L = mC + b\)

\[ \begin{aligned} 124.942 &= 0.002133\overline{3}(20) + b \\ 124.942 &= 0.042666\overline{6} + b \\ b &= 124.899333\overline{3} \end{aligned} \]

Therefore, the required linear relation between \(L\) and \(C\) is

\[ \begin{aligned} L = 0.002133\overline{3}\,C + 124.899333\overline{3} \end{aligned} \]

Hence, the length of the copper rod in terms of temperature is \(L = 0.002133\overline{3}\,C + 124.899333\overline{3}\)

Q16. The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre?

Solution

Let the weekly demand of milk \(D\) (in litres) be a linear function of the selling price \(P\) (in रुपये per litre). Then we can write

\[ \begin{aligned} D = mP + b \end{aligned} \]

From the given information, the two demand–price points are \((14, 980)\) and \((16, 1220)\). The slope \(m\) is

\[ \begin{aligned} m &= \frac{1220 - 980}{16 - 14} \\ &= \frac{240}{2} \\ &= 120 \end{aligned} \]

Substituting \(m = 120\) and the point \((14, 980)\) into \(D = mP + b\)

\[ \begin{aligned} 980 &= 120(14) + b \\ 980 &= 1680 + b \\ b &= -700 \end{aligned} \]

Thus, the linear demand function is

\[ \begin{aligned} D = 120P - 700 \end{aligned} \]

For a selling price of \(Rs\ 17\) per litre,

\[ \begin{aligned} D &= 120(17) - 700 \\ &= 2040 - 700 \\ &= 1340 \end{aligned} \]

Hence, the owner can sell \(1340\) litres of milk per week at \(Rs\ 17\) per litre

Q17. P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is \(\dfrac{x}{a}+\dfrac{y}{b}=2\)

Solution

Let the given line intersect the x-axis at \(A(p, 0)\) and the y-axis at \(B(0, q)\). The equation of a line in intercept form is

\[ \begin{aligned} \frac{x}{p} + \frac{y}{q} = 1 \end{aligned} \]

Since \(P(a, b)\) is the midpoint of the line segment joining \(A(p, 0)\) and \(B(0, q)\), its coordinates are given by the midpoint formula

\[ \begin{aligned} a &= \frac{p + 0}{2} = \frac{p}{2} \\ b &= \frac{0 + q}{2} = \frac{q}{2} \end{aligned} \]

From these relations, we get

\[ \begin{aligned} p &= 2a \\ q &= 2b \end{aligned} \]

Substituting \(p = 2a\) and \(q = 2b\) into the intercept form of the line,

\[ \begin{aligned} \frac{x}{2a} + \frac{y}{2b} &= 1 \\ \frac{x}{a} + \frac{y}{b} &= 2 \end{aligned} \]

Hence, the equation of the line is \(\dfrac{x}{a} + \dfrac{y}{b} = 2\)

Q18. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.

Solution

Let the required line intersect the x-axis at \(A(a, 0)\) and the y-axis at \(B(0, b)\). The equation of the line in intercept form is

\[ \begin{aligned} \frac{x}{a} + \frac{y}{b} = 1 \end{aligned} \]

The point \(R(h, k)\) divides the line segment joining \(A(a, 0)\) and \(B(0, b)\) internally in the ratio \(1 : 2\). Using the section formula, the coordinates of \(R\) are

\[ \begin{aligned} h &= \frac{1 \cdot 0 + 2 \cdot a}{1 + 2} = \frac{2a}{3} \\ k &= \frac{1 \cdot b + 2 \cdot 0}{1 + 2} = \frac{b}{3} \end{aligned} \]

From these relations, we get

\[ \begin{aligned} a &= \frac{3h}{2} \\ b &= 3k \end{aligned} \]

Substituting \(a = \frac{3h}{2}\) and \(b = 3k\) into the intercept form of the line,

\[ \begin{aligned} \frac{x}{\frac{3h}{2}} + \frac{y}{3k} &= 1 \\ \frac{2x}{3h} + \frac{y}{3k} &= 1 \\ \frac{2x}{h} + \frac{y}{k} &= 3 \end{aligned} \]

Hence, the equation of the required line is \(\dfrac{2x}{h} + \dfrac{y}{k} = 3\)

Q19. By using the concept of equation of a line, prove that the three points (3, 0), (– 2, – 2) and (8, 2) are collinear.

Solution

Let the given points be \(A(3, 0)\), \(B(-2, -2)\), and \(C(8, 2)\). To prove that these points are collinear using the concept of the equation of a line, we check whether all three points satisfy a single linear equation.

First, we find the slope of the line joining points \(A(3, 0)\) and \(B(-2, -2)\)

\[ \begin{aligned} m_{AB} &= \frac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{-2 - 0}{-2 - 3} \\ &= \frac{-2}{-5} \\ &= \frac{2}{5} \end{aligned} \]

Now we write the equation of the line passing through \(A(3, 0)\) with slope \(\frac{2}{5}\)

\[ \begin{aligned} y - y_1 &= m(x - x_1) \\ y - 0 &= \frac{2}{5}(x - 3) \\ 5y &= 2x - 6 \\ 2x - 5y - 6 &= 0 \end{aligned} \]

Next, we verify whether point \(C(8, 2)\) satisfies this equation

\[ \begin{aligned} 2(8) - 5(2) - 6 &= 16 - 10 - 6 \\ &= 0 \end{aligned} \]

Since point \(C(8, 2)\) satisfies the same linear equation, it lies on the same straight line as points \(A\) and \(B\)

Hence, the points \((3, 0)\), \((-2, -2)\), and \((8, 2)\) are collinear