CBSE Class 11 Maths Exemplar Chapter 10: Straight Lines NCERT Solutions

Solution:

Let the equation of the straight line be in the intercept form: $\backslash frac\{x\}\{a\}\; +\; \backslash frac\{y\}\{b\}\; =\; 1$, where $a$ and $b$ are the intercepts on the X and Y axes, respectively.

Given that the line cuts off equal intercepts from the axes, we have $a\; =\; b$. Let this common intercept be $k$. So, the equation of the line becomes $\backslash frac\{x\}\{k\}\; +\; \backslash frac\{y\}\{k\}\; =\; 1$.

Multiplying by $k$, we get $x\; +\; y\; =\; k$, or $x\; +\; y\; -\; k\; =\; 0$.

The problem states that the line passes through the point (1, -2). Substituting these coordinates into the equation:

$$1\; +\; (-2)\; -\; k\; =\; 0$$

$$-1\; -\; k\; =\; 0$$

$$k\; =\; -1$$

Now, substitute the value of $k$ back into the equation of the line:

$\backslash frac\{x\}\{-1\}\; +\; \backslash frac\{y\}\{-1\}\; =\; 1$

This simplifies to $-x\; -\; y\; =\; 1$, or $x\; +\; y\; +\; 1\; =\; 0$.

Therefore, the equation of the straight line is $x\; +\; y\; +\; 1\; =\; 0$.

Solution:

First, we need to find the slope of the line joining the points (2, 3) and (3, -1). Let these points be $P(2,\; 3)$ and $Q(3,\; -1)$.

The slope $m\_\{PQ\}$ is given by the formula $m\; =\; \backslash frac\{y\_2\; -\; y\_1\}\{x\_2\; -\; x\_1\}$.

$$m\_\{PQ\}\; =\; \backslash frac\{-1\; -\; 3\}\{3\; -\; 2\}\; =\; \backslash frac\{-4\}\{1\}\; =\; -4$$

The required line is perpendicular to the line PQ. If $m$ is the slope of the required line, then the product of the slopes of two perpendicular lines is -1.

$$m\; \backslash times\; m\_\{PQ\}\; =\; -1$$

$$m\; \backslash times\; (-4)\; =\; -1$$

$$m\; =\; \backslash frac\{-1\}\{-4\}\; =\; \backslash frac\{1\}\{4\}$$

Now, we have the slope of the required line ($m\; =\; \backslash frac\{1\}\{4\}$) and a point it passes through (5, 2).

Using the point-slope form of a line, $y\; -\; y\_1\; =\; m(x\; -\; x\_1)$:

$$y\; -\; 2\; =\; \backslash frac\{1\}\{4\}(x\; -\; 5)$$

Multiply both sides by 4:

$$4(y\; -\; 2)\; =\; 1(x\; -\; 5)$$

$$4y\; -\; 8\; =\; x\; -\; 5$$

Rearrange the terms to get the equation in the standard form:

$$x\; -\; 4y\; -\; 5\; +\; 8\; =\; 0$$

$$x\; -\; 4y\; +\; 3\; =\; 0$$

Thus, the equation of the required line is $x\; -\; 4y\; +\; 3\; =\; 0$.

Solution:

The equations of the two lines are given as:

Line 1: $y\; =\; (2\; -\; \backslash sqrt\{3\})(x+5)$

Line 2: $y\; =\; (2\; +\; \backslash sqrt\{3\})(x-7)$

The slope-intercept form of a line is $y\; =\; mx\; +\; c$, where $m$ is the slope.

From Line 1, the slope $m\_1\; =\; (2\; -\; \backslash sqrt\{3\})$.

From Line 2, the slope $m\_2\; =\; (2\; +\; \backslash sqrt\{3\})$.

Let $\backslash theta$ be the angle between the two lines. The formula for the tangent of the angle between two lines is:

$$\backslash tan\; \backslash theta\; =\; \backslash left|\; \backslash frac\{m\_1\; -\; m\_2\}\{1\; +\; m\_1\; m\_2\}\; \backslash right|$$

Substitute the values of $m\_1$ and $m\_2$:

$$\backslash tan\; \backslash theta\; =\; \backslash left|\; \backslash frac\{(2\; -\; \backslash sqrt\{3\})\; -\; (2\; +\; \backslash sqrt\{3\})\}\{1\; +\; (2\; -\; \backslash sqrt\{3\})(2\; +\; \backslash sqrt\{3\})\}\; \backslash right|$$

Simplify the numerator:

$$2\; -\; \backslash sqrt\{3\}\; -\; 2\; -\; \backslash sqrt\{3\}\; =\; -2\backslash sqrt\{3\}$$

Simplify the denominator using the difference of squares formula $(a-b)(a+b)\; =\; a^2\; -\; b^2$:

$$1\; +\; (2^2\; -\; (\backslash sqrt\{3\})^2)\; =\; 1\; +\; (4\; -\; 3)\; =\; 1\; +\; 1\; =\; 2$$

Now, substitute these back into the $\backslash tan\; \backslash theta$ formula:

$$\backslash tan\; \backslash theta\; =\; \backslash left|\; \backslash frac\{-2\backslash sqrt\{3\}\}\{2\}\; \backslash right|\; =\; |-\backslash sqrt\{3\}|\; =\; \backslash sqrt\{3\}$$

We know that $\backslash tan\; 60^\backslash circ\; =\; \backslash sqrt\{3\}$ or $\backslash tan\; \backslash frac\{\backslash pi\}\{3\}\; =\; \backslash sqrt\{3\}$.

So, the acute angle $\backslash theta\; =\; 60^\backslash circ$ or $\backslash frac\{\backslash pi\}\{3\}$ radians.

The other angle (obtuse angle) between the lines is $180^\backslash circ\; -\; 60^\backslash circ\; =\; 120^\backslash circ$ or $\backslash pi\; -\; \backslash frac\{\backslash pi\}\{3\}\; =\; \backslash frac\{2\backslash pi\}\{3\}$ radians.

Therefore, the angles between the lines are $60^\backslash circ$ and $120^\backslash circ$.

Solution:

Let the equation of the line be in the intercept form: $\backslash frac\{x\}\{a\}\; +\; \backslash frac\{y\}\{b\}\; =\; 1$, where $a$ is the x-intercept and $b$ is the y-intercept.

According to the problem, the sum of the intercepts is 14. So, $a\; +\; b\; =\; 14$.

From this relation, we can express $b$ in terms of $a$: $b\; =\; 14\; -\; a$.

Substitute this into the intercept form of the line equation:

$\backslash frac\{x\}\{a\}\; +\; \backslash frac\{y\}\{14\; -\; a\}\; =\; 1$

We are given that the line passes through the point (3, 4). Substitute $x\; =\; 3$ and $y\; =\; 4$ into the equation:

$$\backslash frac\{3\}\{a\}\; +\; \backslash frac\{4\}\{14\; -\; a\}\; =\; 1$$

To solve for $a$, find a common denominator:

$$\backslash frac\{3(14\; -\; a)\; +\; 4a\}\{a(14\; -\; a)\}\; =\; 1$$

$$\backslash frac\{42\; -\; 3a\; +\; 4a\}\{14a\; -\; a^2\}\; =\; 1$$

$$\backslash frac\{42\; +\; a\}\{14a\; -\; a^2\}\; =\; 1$$

Cross-multiply:

$$42\; +\; a\; =\; 14a\; -\; a^2$$

Rearrange the terms to form a quadratic equation:

$$a^2\; +\; a\; -\; 14a\; +\; 42\; =\; 0$$

$$a^2\; -\; 13a\; +\; 42\; =\; 0$$

Factor the quadratic equation:

$$a^2\; -\; 7a\; -\; 6a\; +\; 42\; =\; 0$$

$$a(a\; -\; 7)\; -\; 6(a\; -\; 7)\; =\; 0$$

$$(a\; -\; 7)(a\; -\; 6)\; =\; 0$$

This gives two possible values for $a$:

$a\; -\; 7\; =\; 0\; \backslash implies\; a\; =\; 7$

$a\; -\; 6\; =\; 0\; \backslash implies\; a\; =\; 6$

Now, find the corresponding values of $b$ using $b\; =\; 14\; -\; a$:

Case 1: If $a\; =\; 7$, then $b\; =\; 14\; -\; 7\; =\; 7$.

The equation of the line is $\backslash frac\{x\}\{7\}\; +\; \backslash frac\{y\}\{7\}\; =\; 1$, which simplifies to $x\; +\; y\; =\; 7$.

Case 2: If $a\; =\; 6$, then $b\; =\; 14\; -\; 6\; =\; 8$.

The equation of the line is $\backslash frac\{x\}\{6\}\; +\; \backslash frac\{y\}\{8\}\; =\; 1$.

To simplify this equation, find a common denominator (24):

$$\backslash frac\{4x\}\{24\}\; +\; \backslash frac\{3y\}\{24\}\; =\; 1$$

$$4x\; +\; 3y\; =\; 24$$

Therefore, the two possible equations for the lines are $x\; +\; y\; =\; 7$ and $4x\; +\; 3y\; =\; 24$.