High School

A line \( l_1 \) has the equation \( 3x + ky - 11 = 0 \). If \( l_1 \) passes through the point \( A(1, 2) \), find:

(a) The value of \( k \) in \( l_1 \). Hence, write the equation of \( l_1 \) in intercept form.

(b) The perpendicular distance of \( l_1 \) from the point \( B(3, 3) \).

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:


  1. 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]


  2. Simplify the equation:

    [tex]3 + 2k - 11 = 0[/tex]


  3. Solve for [tex]k[/tex]:

    [tex]2k = 11 - 3[/tex]

    [tex]2k = 8[/tex]

    [tex]k = \frac{8}{2} = 4[/tex]


  4. 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]:


  1. 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]


  2. 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]


  3. Calculate the numerator:
    [tex]|9 + 12 - 11| = |10| = 10[/tex]


  4. Calculate the denominator:
    [tex]\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5[/tex]


  5. 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.


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