Answer :
To solve the given problem related to the line [tex]l_1[/tex] with the equation [tex]3x + ky - 11 = 0[/tex] that passes through the point [tex]A(1, 2)[/tex], follow these steps:
(a) Finding the value of [tex]k[/tex] and writing the equation in intercept form:
Substitute the point into the equation:
Since [tex]l_1[/tex] passes through the point [tex]A(1, 2)[/tex], substitute [tex]x = 1[/tex] and [tex]y = 2[/tex] into the equation:
[tex]3(1) + k(2) - 11 = 0[/tex]Simplify the equation:
[tex]3 + 2k - 11 = 0[/tex]
Solve for [tex]k[/tex]:
[tex]2k = 11 - 3[/tex]
[tex]2k = 8[/tex]
[tex]k = \frac{8}{2} = 4[/tex]
Write the equation in intercept form:
With [tex]k = 4[/tex], the equation becomes:
[tex]3x + 4y - 11 = 0[/tex]
Rewrite it in intercept form [tex]\frac{x}{a} + \frac{y}{b} = 1[/tex].Rearrange to solve for [tex]y[/tex]:
[tex]4y = -3x + 11[/tex]
Divide through by 4:
[tex]y = -\frac{3}{4}x + \frac{11}{4}[/tex]
Replace [tex]y[/tex] in terms of [tex]x[/tex]:
[tex]\frac{x}{\frac{11}{3}} + \frac{y}{4} = 1[/tex]
(b) Finding the perpendicular distance from [tex]l_1[/tex] to the point [tex]B(3, 3)[/tex]:
Use the distance formula for a point to a line:
The perpendicular distance [tex]d[/tex] from a point [tex](x_1, y_1)[/tex] to the line [tex]Ax + By + C = 0[/tex] is given by:
[tex]d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}[/tex]Substitute the values:
Here, [tex]A = 3[/tex], [tex]B = 4[/tex], [tex]C = -11[/tex], [tex]x_1 = 3[/tex], and [tex]y_1 = 3[/tex].[tex]d = \frac{|3(3) + 4(3) - 11|}{\sqrt{3^2 + 4^2}}[/tex]
Calculate the numerator:
[tex]|9 + 12 - 11| = |10| = 10[/tex]Calculate the denominator:
[tex]\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5[/tex]Find the distance:
[tex]d = \frac{10}{5} = 2[/tex]So, the perpendicular distance from point [tex]B(3, 3)[/tex] to line [tex]l_1[/tex] is 2 units.