Question

#Quality Question
@ Coordinate Geometry


The equation of the straight line passing through the point (4 , 3) and making intercepts on the co-ordinate Axes whose sum is -1 is :-​

Answer 1

$༒ Question ➽</p><p></p><p>If θ is Acute angle and 3sinθ = 4cosθ then Find out the value of 4sin2θ - 3cos2θ + 2.</p><p></p><p>___________________________</p><p></p><p>༒ Given ➽</p><p></p><p>\large \sf3sin \theta = 4cos\theta3sinθ=4cosθ</p><p></p><p>___________________________</p><p></p><p>༒ To Find ➽</p><p></p><p>\sf \large 4sin2\theta - 3cos2\theta + 24sin2θ−3cos2θ+2</p><p></p><p>___________________________</p><p></p><p>༒ Solution ➽</p><p></p><p>As</p><p></p><p>\begin{gathered} \sf3sin\theta = 4cos\theta \\ \\ \implies \sf \dfrac{sin\theta }{cos \theta } = \frac{4}{3} \\ \\ \implies \large \boxed{ \bf tan\theta = \frac{4}{3} }\end{gathered}3sinθ=4cosθ⟹cosθsinθ=34⟹tanθ=34</p><p></p><p>So,</p><p></p><p>\begin{gathered} \sf {tan}^{2} \theta = \frac{16}{9} \\ \\ \sf \implies {sec}^{2} \theta = 1 + \frac{16}{9} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \{ \because\bf {sec}^{2} \theta = 1 + tan {}^{2} \theta \} \\ \\ = \sf \frac{25}{9} \\ \\ \implies \sf sec \theta = \pm \frac{5}{3} \\ \ \bf \{but \: sec\theta \ne - ve \} \: \: ( \because \theta \: is \: acute) \\ \\ \therefore \bf sec\theta = \frac{5}{3} \end{gathered}tan2θ=916⟹sec2θ=1+916{∵sec2θ=1+tan2θ}=925⟹secθ=±35 {butsecθ=−ve}(∵θisacute)∴secθ=35</p><p></p><p>Hence,</p><p></p><p>\begin{gathered} \purple{ \large \underline{ \boxed{\bf cos\theta = \dfrac{3}{5}}}} \\ \: \: \: \: \: \: \: \{ \bf\because cos\theta = \frac{1}{sec \theta } \}\end{gathered}cosθ=53{∵cosθ=secθ1}</p><p></p><p>So,</p><p></p><p>\begin{gathered} \sf sin\theta = \sqrt{1 - {cos}^{2} \theta } \\ \\ \sf = \sqrt{1 - \frac{9}{25} } \\ \\ = \sf \sqrt{ \frac{16}{25} } \\ \\ \sf = \pm\frac{4}{5} \end{gathered}sinθ=1−cos2θ=1−259=2516=±54</p><p></p><p>As θ is Acute so sinθ will not negative</p><p></p><p>Hence,</p><p></p><p>\large \purple{ \bf \underline{ \boxed{ \bf sin \theta = \frac{4}{5} }}}sinθ=54</p><p></p><p>Now,</p><p></p><p>\begin{gathered} \sf4sin2\theta - 3cos2\theta + 2 \\ \\ = \sf 4(2sin\theta cos\theta) - 3(2 {cos}^{2} \theta - 1) + 2 \\ \\ \sf = 8 \times \frac{4}{5} \times \frac{3}{5} -3 \{2( \frac{3}{5} ) {}^{2} - 1 \} + 2 \\ \\ = \sf \frac{96}{25} + \frac{21}{25} + 2 \\ \\ \bf = \frac{167}{25} \end{gathered}4sin2θ−3cos2θ+2=4(2sinθcosθ)−3(2cos2θ−1)+2=8×54×53−3{2(53)2−1}+2=2596+2521+2=25167</p><p></p><p>\begin{gathered} \Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}\end{gathered} Which  is  the  required Answer.</p><p></p><p></p><p>$

༒ Question ➽

If θ is Acute angle and 3sinθ = 4cosθ then Find out the value of 4sin2θ - 3cos2θ + 2.

___________________________

༒ Given ➽

\large \sf3sin \theta = 4cos\theta3sinθ=4cosθ

___________________________

༒ To Find ➽

\sf \large 4sin2\theta - 3cos2\theta + 24sin2θ−3cos2θ+2

___________________________

༒ Solution ➽

As

\begin{gathered} \sf3sin\theta = 4cos\theta \\ \\ \implies \sf \dfrac{sin\theta }{cos \theta } = \frac{4}{3} \\ \\ \implies \large \boxed{ \bf tan\theta = \frac{4}{3} }\end{gathered}

3sinθ=4cosθ

⟹

cosθ

sinθ

=

3

4

⟹

tanθ=

3

4

So,

\begin{gathered} \sf {tan}^{2} \theta = \frac{16}{9} \\ \\ \sf \implies {sec}^{2} \theta = 1 + \frac{16}{9} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \{ \because\bf {sec}^{2} \theta = 1 + tan {}^{2} \theta \} \\ \\ = \sf \frac{25}{9} \\ \\ \implies \sf sec \theta = \pm \frac{5}{3} \\ \ \bf \{but \: sec\theta \ne - ve \} \: \: ( \because \theta \: is \: acute) \\ \\ \therefore \bf sec\theta = \frac{5}{3} \end{gathered}

tan

2

θ=

9

16

⟹sec

2

θ=1+

9

16

{∵sec

2

θ=1+tan

2

θ}

=

9

25

⟹secθ=±

3

5

{butsecθ



=−ve}(∵θisacute)

∴secθ=

3

5

Hence,

\begin{gathered} \purple{ \large \underline{ \boxed{\bf cos\theta = \dfrac{3}{5}}}} \\ \: \: \: \: \: \: \: \{ \bf\because cos\theta = \frac{1}{sec \theta } \}\end{gathered}

cosθ=

5

3

{∵cosθ=

secθ

1

}

So,

\begin{gathered} \sf sin\theta = \sqrt{1 - {cos}^{2} \theta } \\ \\ \sf = \sqrt{1 - \frac{9}{25} } \\ \\ = \sf \sqrt{ \frac{16}{25} } \\ \\ \sf = \pm\frac{4}{5} \end{gathered}

sinθ=

1−cos

2

θ

=

1−

25

9

=

25

16

=±

5

4

As θ is Acute so sinθ will not negative

Hence,

\large \purple{ \bf \underline{ \boxed{ \bf sin \theta = \frac{4}{5} }}}

sinθ=

5

4

Now,

\begin{gathered} \sf4sin2\theta - 3cos2\theta + 2 \\ \\ = \sf 4(2sin\theta cos\theta) - 3(2 {cos}^{2} \theta - 1) + 2 \\ \\ \sf = 8 \times \frac{4}{5} \times \frac{3}{5} -3 \{2( \frac{3}{5} ) {}^{2} - 1 \} + 2 \\ \\ = \sf \frac{96}{25} + \frac{21}{25} + 2 \\ \\ \bf = \frac{167}{25} \end{gathered}

4sin2θ−3cos2θ+2

=4(2sinθcosθ)−3(2cos

2

θ−1)+2

=8×

5

4

×

5

3

−3{2(

5

3

)

2

−1}+2

=

25

96

+

25

21

+2

=

25

167

\begin{gathered} \Large \red{\mathfrak{ \text{W}hich \: \: is \: \: the \: \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}\end{gathered}

Which is the required

Answer.

Answer 2

$\large\underline{\sf{Given- }}$

The straight line passing through the point (4 , 3) and making intercepts on the co-ordinate axes whose sum is -1.

$\large\underline{\sf{To\:Find - }}$

The equation of line.

$\large\underline{\sf{Solution-}}$

Let assume that the line makes an intercept of 'a' units on x - axis and intercept of 'b' units on y - axis.

Then equation of line in intercept form is given by

$\rm :\longmapsto\:\dfrac{x}{a} + \dfrac{y}{b} = 1 - - - (1)$

Now,

According to statement,

It is given that line making intercept on the axis whose sum is - 1.

$\rm :\longmapsto\:a + b = - 1$

So,

$\rm :\longmapsto\:b = - 1 - a - - - (2)$

Substituting the value of b in equation (1), we get

$\rm :\longmapsto\:\dfrac{x}{a} + \dfrac{y}{ - 1 - a} = 1$

$\rm :\longmapsto\:\dfrac{x}{a} - \dfrac{y}{ 1 + a} = 1 - - - (3)$

According to statement,

It is given that line (3) passes through the point ( 4, 3 ).

So,

$\rm :\longmapsto\:\dfrac{4}{a} - \dfrac{3}{ 1 + a} = 1$

$\rm :\longmapsto\:\dfrac{4(1 + a) - 3a}{a(1 + a)} = 1$

$\rm :\longmapsto\:\dfrac{4 + 4a- 3a}{a(1 + a)} = 1$

$\rm :\longmapsto\:\dfrac{4 + a}{a(1 + a)} = 1$

$\rm :\longmapsto\: {a}^{2} + a = 4 + a$

$\rm :\longmapsto\: {a}^{2}= 4$

$\bf\implies \:a \: = \: \pm \: 2$

So, required equation is obtained by substituting a = 2 and a = - 2 in equation (3),

$\bf :\longmapsto\:\dfrac{x}{2} - \dfrac{y}{3} = 1 \implies \: 3x - 2y = 6$

or

$\bf :\longmapsto\:\dfrac{x}{ - 2} + \dfrac{y}{1} = 1 \implies \: - x + 2y = 2$

Additional Information :-

Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

Equation of the line passes through h, k) which is parallel to the x-axis is y = k.

and

Equation of line which is parallel to y-axis is x = h

2. Point-slope form

Consider a line whose slope is m and passes through the point ( a, b ), then equation of line is given by y - b = m(x - a)

3. Slope-intercept form

Consider a line whose slope is m which cuts the y-axis at a distance ‘a’ from the origin then equation of line is given by y = mx + a.

4. Intercept Form of Line

Consider a line having x– intercept a and y– intercept b, then the equation of line is x/a + y/b = 1.